THE  LIBRARY 

OF 

THE  UNIVERSITY 
OF  CALIFORNIA 

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GIFT  OF 

John  S.Prell 


MATHEMATICAL   WORKS, 
BY   PROFESSOR   EDWARD   A.    BOWSER. 

ACADEMIC  ALGEBRA.    With  Numerous  Examples. 
COLLEGE  ALGEBRA.    With  Numerous  Examples. 

PLANE  AND  SOLID  GEOMETRY.  With  Numerous 
Exercises. 

ELEMENTS  OF  PLANE  AND  SPHERICAL  TRIG- 
ONOMETRY. With  Numerous  Examples. 

A  TREATISE  ON  PLANE  AND  SPHERICAL  TRIG- 
ONOMETRY, AND  ITS  APPLICATIONS  TO  ASTRONOMY 
AND  GEODESY.  With  Numerous  Examples. 

AN  ELEMENTARY  TREATISE  ON  ANALYTIC 
GEOMETRY,  EMBRACING  PLANE  GEOMETRY  AND 
AN  INTRODUCTION  TO  GEOMETRY  OF  THREE  DIMEN- 
SIONS. With  Numerous  Examples. 

AN  ELEMENTARY  TREATISE  ON  THE  DIFFER- 
ENTIAL AND  INTEGRAL  CALCULUS.  With 
Numerous  Examples. 

AN  ELEMENTARY  TREATISE  ON  ANALYTIC 
MECHANICS.  With  Numerous  Examples. 

AN  ELEMENTARY  TREATISE  ON  HYDROME- 
CHANICS. With  Numerous  Examples. 

LOGARITHMIC  AND  TRIGONOMETRIC  TABLES. 

A  TREATISE  ON  ROOFS  AND  BRIDGES.  With 
Numerous  Exercises. 


JOHfl  S.  PRELL 

Gvil  &  Mechanical  Engineer, 

SAN  FRANCISCO,  CAL. 
A  TREATISE 


ROOFS  AND   BRIDGES 


WITH  NUMEROUS  EXERCISES 


BY 

EDWARD   A.   BOWSER 

PROFESSOR  OF  MATHEMATICS  AND    ENGINEERING   IN    RUTGERS   COLLEGE 

SECOND   EDITION 


El  646967 


NEW  YORK 
D.    VAN    NOSTRAND    COMPANY 

23  MURRAY  AND  27  WARREN  STREETS 
1902 


COPTBIGHT,  1898, 

BY  E.  A.  BOWBEE. 


NorfoooB 
J.  a  CMhing  &  Co.  -  Berwick  &  Smith 
Norwood  Mass.  U.S.A. 


EigimeriBg 


PREFACE. 

THE  present  treatise  on  Roofs  and  Bridges  is  designed  as 
a  text-book  for  the  use  of  schools.  The  object  of  this  work 
is  to  develop  the  principles  and  explain  the  methods  em- 
ployed in  finding  the  forces  in  Roofs  and  Bridges,  and  to 
train  the  student  to  compute  the  stresses,  due  to  the  dead, 
live,  snow,  and  wind  loads,  in  the  different  members  of  any 
of  the  simple  roof  and  bridge  trusses  that  are  in  common 
use. 

The  aim  has  been  to  explain  the  principles  clearly  and 

concisely,   to   develop   the    different   methods   simply   and 

neatly,  and  to  present  the  subject  in  accordance  with  the 

>  methods  used  in  the  modern  practice  of  roof  and  bridge 

-,T    construction. 

^\      In  introducing  each  new  truss  it  is  at  first  carefully  de- 
scribed, and  the  method  of  loading  it  explained.     A  prob- 
V  lem  is  then  given  for  this  truss,  and  solved  to  determine  the 
^  stresses  in  all  the  members.     This  problem  is  followed  by 
several  other  similar  ones,  which  are  to  be  solved  by  the 
student.     Nearly  all  of  these  problems  were  prepared  espe- 
_  cially  for  this  work,  and  solved  to  obtain  the  answers.     The 
^*  -instructor  can  at  any  time  easily  make  up  problems  for  his 
-  pupils  without  the  answers. 

The  book  consists  of  four  chapters.     Chapter  I.  is  entirely 
~^"  given  to  Roof  Trusses.     Chapter  II.  treats  only  of  Bridge 
Trusses  with  Uniform  Loads, 
iii 


733586 


iv  PREFACE. 

Chapter  III  is  devoted  to  Bridge  Trusses  with  Unequal 
Distribution  of  the  Loads.  This  is  divided  into  three  parts, 
as  follows : 

(1)  The  use  of  a  uniformly  distributed  excess  load  covering 
one  or  more  panels,  followed  by  a  uniform  train  load  cover- 
ing the  whole  span. 

(2)  The  use  of  one  or  two  concentrated  excess  loads,  with 
a  uniform  train  load  covering  the  span. 

(3)  The  use  of  the  actual  specified  locomotive  wheel  loads, 
followed  by  a  uniform  train  load. 

Chapter  IV  treats  of  Miscellaneous  Trusses,  including  the 
Crescent  Roof  Truss,  the  Pegram  and  Parabolic  Bowstring 
Bridge  Trusses,  and  Skew  Bridges. 

The  stresses  in  this  work  are  nearly  all  given  in  tons,  the 
word  "  ton  "  meaning  a  ton  of  2000  pounds.  Any  other 
unit  of  load  and  of  stress  might  be  used  as  well. 

My  best  thanks  are  due  to  my  friend  and  former  pupil, 
Mr.  George  H.  Blakeley,  C.E.,  of  the  class  of  '84,  now 
Chief  Engineer  of  the  Passaic  Rolling  Mill  Company,  for 
reading  the  manuscript  and  for  valuable  suggestions. 


A.  B. 


RUTGERS  COLLEGE, 
NEW  BRUNSWICK,  N.J.,  October,  1898. 


TABLE  OF  CONTENTS. 

CHAPTER  I. 
ROOF  TRUSSES. 

ART.  PAGE 

1.  Definitions         '.        .        .     •  '. 1 

2.  The  Dead  Load         '.                        3 

3.  The  Live  Load .........  4 

4.  The  Apex  Loads  and  Reactions       .....  4 

5.  Relations  between  External  Forces  and  Internal  Stresses  6 

6.  Methods  of  Calculation 11 

7.  Lever  Arms  —  Indeterminate  Cases         ....  14 

8.  Snow  Load  Stresses  .         .         .        .        .        .        .        .19 

9.  Wind' Loads 20 

10.  Wind  Apex  Loads  and  Reactions    .        .        .        .        .  22 

11.  Wind  Stresses .25 

12.  Complete  Calculation  of  a  Roof  Truss     ....  29 

CHAPTER  II. 
BRIDGE  TRUSSES  WITH  UNIFORM  LOADS. 

13.  Definitions .        .38 

14.  Different  Forms  of  Trusses 39 

15.  The  Dead  Load 42 

16.  The  Live  Load 44 

17.  Shear  — Shearing  Stress 46 

18.  Web  Stresses  due  to  Dead  Loads      .....  47 

19.  Chord  Stresses  due  to  Dead  Loads  .  50 


vi  CONTENTS. 

AKT.  PAGE 

20.  Position   of    Uniform  Live    Load    causing    Maximum 

Chord  Stresses 54 

21.  Maximum  Stresses  in  the  Chords 55 

22.  Position  of  Uniform  Live  Load  causing  Maximum  Shears  57 

23.  The  Warren  Truss 60 

24.  Mains  and  Counters 70 

25.  The  Howe  Truss 74 

26.  The  Pratt  Truss 81 

27.  The  Warren  Truss  with  Vertical  Suspenders  ...  84 

28.  The  Double  Warren  Truss 87 

29.  The  Whipple  Truss 90 

30.  The  Lattice  Truss 94 

31.  The  Post  Truss 98 

32.  The  Bollman  Truss 100 

33.  The  Fink  Trass 101 

34.  The  Parabolic  Bowstring  Truss 103 

35.  The  Circular  Bowstring  Truss 108 

30.     Snow  Load  Stresses 109 

37.  Stresses  due  to  Wind  Pressure 110 

38.  The  Factor  of  Safety        .        : 118 

CHAPTER  HI. 

BRIDGE  TRUSSES  WITH  UNEQUAL  DISTRIBUTION 
OF  THE  LOADS. 

89.     Preliminary  Statement 120 

40.  When  the  Uniform  Train  Load  is  preceded  by  One  or 

More  Heavy  Excess  Panel  Loads       ....  120 

41.  When  One  Concentrated  Excess  Load  accompanies  a 

Uniform  Train  Load  .        ....        .        .  127 

42.  When  Two  Equal  Concentrated  Excess  Loads  accompany 

a  Uniform  Train  Load        .         .         .         .         .         .  133 

43.  The  Baltimore  Truss 142 

44.  True  Maximum  Shears  for  Uniform  Live  Load       .         .  154 

45.  Locomotive  Wheel  Loads                                                     .  157 


CONTENTS.  vii 

46.  Position  of  Wheel  Loads  for  Maximum  Shear         .        .     158 

47.  Position  of   Wheel   Loads  for  Maximum  Moment  at 

Joint  in  Loaded  Chord       .         .        .        .         .         .164 

48.  Position  of  Wheel   Loads  for   Maximum   Moment  at 

Joint  in  Unloaded  Chord 166 

49.  Tabulation  of  Moments  of  Wheel  Loads.        .        .        .168 


CHAPTER  IV. 

MISCELLANEOUS  TRUSSES. 

Roof  Trusses. 

50.  The  King  and  Queen  Truss  —  The  Fink  Truss       .        .     177 

51.  The  Crescent  Truss  .        .        .        .        .        .  .    182 

Bridge  Trusses. 

52.  The  Pegram  Truss  —  The  Parabolic  Bowstring  Truss    .     187 
63     Skew  Bridges 192 


ROOFS   AND   BRIDGES. 


CHAPTER  I. 

ROOF  TRUSSES. 

Art.   1.    Definitions.  —  A  Framed  Structure  is  a 

collection  of  pieces,  either  of  wood  or  metal,  or  both 
combined,  so  joined  together  as  to  cause  the  structure  to 
act  as  one  rigid  body.  The  points  at  which  the  pieces  are 
joined  together  are  usually  called  joints. 

A  Truss  is  a  structure  designed  to  transfer  loads  on  it 
to  the  supports  at  each  end,  while  each  member  of  the  truss 
is  subject  only  to  longitudinal  stress,  either  tension  or  com- 
pression. The  simplest  of  all  trusses  is  a  triangle ;  and  all 
trusses,  however  complicated,  containing  no  superfluous 
members,  must  be  composed  of  an  assemblage  of  triangles, 
since  a  triangle  is  the  only  polygon  whose  form  cannot  be 
changed  without  changing  the  lengths  of  its  sides. 

A  Strut  is  a  member  which  takes  compression.  Struts 
are  sometimes  called  posts,  or  columns. 

A  Tie  is  a  member  which  takes  tension. 

A  Brace  is  a  term  used  to  denote  both  struts  and  ties. 

A  Counter  Brace  is  a  member  which  is  designed  to  take 
both  compression  and  tension.  A  Counter  is  a  member 
designed  to  take  either  compression  or  tension ;  that  is,  for 
one  position  of  the  load  the  member  may  be  compressed, 
while  for  another  position  it  may  be  elongated. 
1 


2  ROOFS  AND  BRIDGES. 

The  Upper  and  Lower  Chords  are  the  upper  and 
lower  members  of  a  truss,  extending  from  one  support  to 
the  other.  Each  half  of  the  upper  chord  of  a  roof  truss 
is  sometimes  called  the  "main  rafter,"  while  the  lower 
chord  is  often  called  the  "tie  rod."  The  upper  chord  is 
always  in  compression  and  the  lower  chord  is  always  in 
tension.  The  spaces  between  the  joints  of  the  chords  are 
called  panels. 

The  Web  Members  are  those  which  connect  the  joints 
of  the  two  chords.  They  are  generally  alternately  struts 
and  ties. 

A  Roof,  in  common  language,  is  the  covering  over  a 
structure,  the  chief  object  of  which  is  to  protect  the  build- 
ing from  the  effects  of  rain  and  snow. 

A  Roof  Truss  is  a  structure  which  supports  a  roof. 
Eoof  Trusses  are  of  almost  innumerable  forms,  and  they 
differ  greatly  in  the  details  of  their  construction. 

The  External  Forces  include  all  the  exterior  or  applied 
forces,  such  as  the  weight  of  the  structure,  the  weight  of 
snow,  the  force  of  the  wind,  the  reactions,  etc.,  which  act 
upon  and  tend  to  distort  the  structure.  The  external  forces 
on  the  whole  structure  must  balance  each  other,  or  else  the 
whole  structure  will  begin  to  move. 

Strain  is  a  change  in  the  length  or  in  the  form  of  a  body, 
which  has  been  produced  by  the  application  of  one  or  more 
external  forces ;  and  it  is  to  be  measured,  not  in  tons,  but  in 
units  of  length,  as  inches  or  feet. 

Stress  is  the  name  given  to  that  internal  force  which  is 
exerted  by  the  material  in  resisting  strain ;  and  it  is  meas- 
ured in  pounds  or  tons  the  same  as  the  external  forces.  It 
follows  that,  when  every  part  of  a  strained  member  of  a 
structure  is  in  equilibrium,  the  internal  stress  exerted  at 
any  imaginary  section  through  the  member  is  equal  and 
opposite  to  the  straining  force. 


ROOF  TRUSSES.  6 

Various  Kinds  of  Stresses.  —  The  external  forces  may 
produce,  according  to  circumstances,  different  internal  forces 
or  stresses  in  the  various  pieces  of  the  structure.  These 
forces  or  stresses  and  their  accompanying  strains  may  be 
classified  as  follows  : 

1.  A  direct  pull  or  a  tensile  stress,  producing  extension 
or  elongation. 

2.  A  direct  thrust  or  a  compressive   stress,   producing 
compression. 

3.  A  shearing  force  or  a  shearing   stress,  producing  a 
cutting  asunder. 

A  shearing  force  or  shearing  stress  is  caused  by  two  forces 
acting  parallel  to  each  other,  and  at  right  angles  to  the  axis 
of  the  piece,  but  in  opposite  directions,  and  in  two  imme- 
diately consecutive  sections.  The  tendency  is  to  cause  two 
adjacent  sections  to  slide  on  each  other;  and  the  term 
shearing  force,  or  shearing  stress,  is  used,  because  the 
force  is  similar  to  that  in  the  blades  of  a  pair  of  shears  in 
the  act  of  cutting. 

Art.  2.  The  Dead  Load.  —  The  dead  or  fixed  load  sup- 
ported by  a  roof  truss  consists  of  the  weight  of  the  truss 
itself  and  the  weight  of  the  roof.  The  weight  of  the  truss 
can  only  be  approximated.  The  following  formula*  gives 
approximately  correct  results  for  short  spans  :  Let  I  —  span 
in  feet,  b  =  distance  between  trusses  in  feet;  and  W=  ap- 
proximate weight  of  one  truss  in  pounds.  Then 


The  roof  includes  the  roof-covering,  the  sheeting,  the 
rafters,  and  the  purlins.  Its  total  weight  will  vary  from 
5  to  30  Ibs.  per  square  foot  of  roof  surface.  The  purlins  are 

*  Modern  Framed  Structures,  by  Johnson,  Bryan,  and  Turneaure. 


4  ROOFS  AND  BRIDGES. 

beams  running  longitudinally  between  the  trusses,  and  fast- 
ened to  them  at  the  upper  joints,  and  often  midway  between 
them  as  well. 

Art.  3.  The  Live  Load.  —  The  live  or  variable  load 
consists  of  the  snow  load  and  the  wind  load. 

The  snow  load  varies,  according  to  the  locality,  from  10  Ibs. 
to  30  Ibs.  per  square  foot  of  horizontal  projection.  The 
weight  of  new  snow  varies  from  5  Ibs.  to  12  Ibs.  per  cubic 
foot,  although  snow  which  is  saturated  with  water  weighs 
much  more.  The  snow  load  need  not  be  considered  when 
the  inclination  of  the  roof  to  the  horizontal  is  60°  or  more, 
as  the  snow  would  slide  off. 

The  wind  load  is  variable  in  direction  and  intensity,  and 
often  injurious  in  its  effects ;  especially  is  this  the  case  with 
large  trusses  placed  at  considerable  intervals  apart.  The 
maximum  wind  pressure  on  a  surface  normal  to  its  direction 
is  variously  estimated  at  from  30  Ibs.  to  56  Ibs.  per  square 
foot.  The  calculation,  therefore,  of  the  stresses  caused  by 
the  wind  forces  is  often  of  considerable  importance,  and 
should  never  be  left  out  of  account  in  designing  iron  roofs 
of  large  span. 

Art.  4.  The  Apex  Loads  and  Reactions.  —  Both  the 
dead  and  live  loads  are  transferred,  by  means  of  the  purlins, 
to  the  joints  or  apexes  of  the  upper  chords  of  the  truss ;  and 
these  loads  at  the  apexes  are  called  the  apex  loads.  They 
are  also  called  the  panel  loads. 

The  truss,  roof,  and  snow  loads,  being  vertical,  and  uni- 
formly distributed  over  each  panel,  the  apex  loads  are 
each  equal  to  one  half  the  sum  of  the  adjacent  panel  loads. 
Thus,  the  load  at  h,  Fig.  1,  is  equal  to  one  half  the  panel 
load  on  he  plus  one  half  the  panel  load  on  ah.  The  wind 


ROOF  TRUSSES. 


load  at  h  is  also  equaj  to  one  half  the  wind  load  on  he  plus 
one  half  the  wind  load  on  ah,  the  load  on  each  being  normal 
to  the  siirface. 

If  the  truss  be  symmetri- 
cal, and  the  panels  be  of  equal 
length,  the  apex  loads  are 
determined  by  dividing  the 
total  load  by  the  number  of 
panels  in  the  upper  chords. 
Thus  in  Fig.  1,  the  apex  loads  ri°'  * 

at  h  and  c  are  each  one  fourth  of  the  total  load.  At  the  sup- 
ports, a  and  b,  the  loads  are  only  one  half  those  at  h  and  c. 

Reactions.  —  For  dead  and  snow  loads  both  reactions  of 
the  supports  of  the  truss  are  vertical,  and  each  is  one  half 
of  the  total  load,  if  the  truss  is  symmetrical.  For  wind  load 
the  reactions  depend  upon  the  manner  of  supporting  the 
truss. 

Prob.  1.  A  roof  truss,  like  Fig.  1,  has  its  span  80  feet  and 
its  rise  40  feet;  the  distance  between  trusses  is  12  feet, 
center  to  center.  Find  (1)  the  weight  of  the  truss,  (2)  the 
weight  of  the  roof,  (3)  the  snow  load,  (4)  the  apex  loads, 
and  (5)  the  reactions. 

For  these  problems,  take  20  Ibs.  per  square  foot  of  roof  surface  for 
•weight  of  roof,  and  20  Ibs.  per  square  foot  of  horizontal  projection  for 
snow  load. 


Here  ab  =  80  feet,  dc  =  40  feet,  and  ac  =  56.56  feet. 
Weight  of  truss,  Wt  =  ^  bl2  =    3200 


Weight  of  roof,  Wr  =  56.56  x  20  x  12  x  2  =  27148.8  Ibs. 

Weight  of  snow,  W.  =  80  x  12  x  20             =  19200     Ibs. 

Each  apex  load  =  \  x  49548.8               =  12387.2  Ibs. 

Each  reaction,  R  =  |  x  49548.8                =  24774.4  Ibs. 


Ibs. 


ROOFS  AND  BRIDGES. 


Prob.  2.   A  roof  truss,  like  Fig.  2,  has  its  span  90  feet,  its 
rise  30  feet,  and  distance  between  trusses  13  feet:  find  the 


Trig. 


total  apex  loads  and  the  reactions  for  the  weights  of  the 
truss,  roof,  and  snow. 

Ans.  Apex  loads  =  6989  and  13977  Ibs.  ;  reaction  = 
27954  Ibs. 

Prob.  3.  In  the  roof  truss  of  Fig.  3,  the  span  is  100  feet, 
the  rise  is  25  feet,  and  the  distance  between  trusses  12.5 
feet:  find  the  total  apex  loads  and  the  reactions  for  the 
truss,  roof,  and  snow  loads. 


a 

Fig.  3 

Ans.  Loads  =  7270  and  14540  Ibs.;  reaction  =  29079  Ibs. 

Art.  5.  Relations  between  External  Forces  and 
Internal  Stresses.  —  The  external  forces  acting  upon  a 
truss  are  in  equilibrium  with  the  internal  stresses  in  the 
members  of  the  truss. 

Let  ran  be  a  section  passed  through  the  truss,  Fig.  4,  cut- 


HOOF  THUSSES.  1 

ting  the  members  whose  stresses  are  desired,  and  let  these 
stresses  be  replaced  by  equal  external  forces.     Then  it  is 


clear  that  the  equilibrium  is  undisturbed.  Therefore  we 
have  the  principle : 

The  internal  stresses  in  any  section  hold  in  equilibrium 
the  external  forces  acting  upon  either  side  of  that  section. 

If  we  remove  either  portion  of  the  truss,  as  the  one  on 
the  right  of  the  section,  then  the  external  forces  on  the 
remaining  part  of  the  truss,  together  with  the  internal 
stresses,  form  a  system  of  forces  in  static  equilibrium. 
And  from  Anal.  Mechanics,  Art.  61,  the  conditions  of 
equilibrium  for  a  system  of  forces  acting  in  any  direction 
in  one  plane  on  a  rigid  body  are: 

2  horizontal  components  =0 (1) 

2  vertical  components      =0 (2) 

2  moments  =  0   .     .     .     .  •  .     .     .     .     (3) 

These  three  equations  of  condition  state  the  relations, 
between  the  internal  stresses  in  any  section,  and  the  ex- 
ternal forces  on  either  side  of  that  section.  If  only  three 
of  these  internal  stresses  are  unknown,  they  can  therefore 
be  determined. 

For  example,  in  Fig.  4,  let  jR,  P0,  Pl  be  the  reaction  and 
apex  loads,  found  as  in  Art.  4;  let  sr,  s2,  s3  be  the  stresses 
in  the  members  he,  Jcc,  and  M,  that  are  cut  by  the  section 
mn,  and  let  0j  and  02  be  the  angles  made  by  Sj  and  s2  with 


8  ROOFS  AND  BRIDGES. 

the  vertical.     Applying  our  three  equations  of  equilibrium, 

we  have : 

for  horizontal  components,  Sj  sin  Ol  +  s2  sin  02  +  s3  =  0, 
for  vertical  components,  R — P0— PI  +  Si  cos  0X + s2  cos  0  2 = 0. 
In  applying  our  third  equation  of  condition,  the  center  of 

moments   may  be   chosen  at  any  point,  Anal.  Mechanics, 

Art.  46.     If  we  take  it  at  c,  the  moments  of  sl  and  s2  are 

zero,  and  the  equation  is  — 


These  three  equations  enable  us  to  find  the  unknown 
stresses. 

NOTE  1.  —  In  all  cases,  in  this  work,  a  tensile  stress  is  denoted  by 
the  positive  sign,  and  a  conipressive  stress,  by  the  negative  sign.*  In 
stating  the  equations,  it  will  be  convenient  to  represent  the  unknown 
stresses  as  tensile,  pulling  away  from  the  section,  as  in  Fig.  4.  Then, 
if  the  numerical  values  of  these  stresses  are  found  to  be  positive,  it 
will  show  that  they  were  assumed  in  the  right  direction,  and  are  ten- 
sile ;  but  if  they  are  negative,  they  were  assumed  in  the  wrong  direc- 
tion, and  are  compressive. 

Prob.  4.  In  the  truss  of  Fig.  5  the  span  is  90-feet,  the 
rise  is  35  feet,  hk  is  perpendicular  to  the  rafter  at  its  mid- 
point, and  the  loads  are  as  shown :  let  it  be  required  to  find 
the  stresses  in  all  the  members  of  the  truss. 

Representing  the  stresses  by  sl}  s2,  ss,  s4,  s5,  and  s6,  we 
proceed  to  apply  our  three  equations  of  equilibrium. 

To  find  the  stress  s^  pass  a  section  cutting  Si  and  ss,  sepa- 
rating the  portion  to  the  left,  take  moments  about  7t,  and 
regard  st  as  pulling  toward  the  right  from  the  section 
(see  Note  1).  Then,  2  moments  about  h  =  0  gives 

(20000  -  5000)  22£  -8lx  17|  =  0.     .:  Sl  =  19287  Ibs. 

*  This  convention  is  entirely  arbitrary.  Some  writers  denote  com- 
pression by  the  positive  sign,  and  tension  by  the  negative  sign. 


EOOF  TRUSSES. 


To  find  s2,  pass  a  section  cutting  *2,  s6,  and  s4,  take  mo- 
ments about  c,  and  regard  s2  as  pulling  toward  the  right. 


Fig.  5 

2  mom.  about  c  =  0  gives 

(20000  -  5000)  45  -  10000  x  22£  -  s2  x  35  =  0. 
.-.  s2  =  12857  Ibs. 

To  find  ss,  pass  a  section  cutting  Sj  and  s3 ;  we  might  take 
moments  about  Jc,  but  we  will  use  our  second  equation 
instead. 

2  vert.  comp.  =  0  gives 

20000  -  5000  +  s3  sin  cad  =  0, 

or,          15000  +  M  «s  =  0  ("since  sin  cad  =  —  =  — 
y  CLC     57 

.-.  s3  =  —  24428  Ibs.,  that  is,  compression  (see  Note  1). 

To  find  s4,  pass  a  section  cutting  s4,  se,  and  s2,  take  mo- 
ments about  k,  and  find  the  lever  arms. 
2  mom.  about  A:  =  0  gives 

(20000  -  5000)36.1  -  10000  x  13.6  +  s4  x  22.17  =  0 
(since  ak  =  ah  sec  cad  =  36.1  and  Jik  =  ah  tan  cad  ==  22.17). 
.-.  s4  =  —  18290  Ibs.,  that  is,  compression. 


10  EOOFS  AND   BRIDGES. 

To  find  s5,  pass  the  section  cutting  s1}  st,  and  s4,  and  take 
moments  about  a.  Then 

10000  x  4£  +  s5  x  4f  =  0.     .-.  s5  =  -  7895  Ibs. 

To  find  s6,  pass  the  section  cutting  s2,  s6,  and  s4,  and  take 
moments  about  a.  Then, 

10000  x  22.5 -s6x  35=0  (since  the  lever  arm  of  se=cd=35), 
.-.  s6  =  6428  Ibs. 

Since  the  truss  is  symmetrical  and  symmetrically  loaded, 
it  is  evident  that  the  stresses  in  all  the  pieces  of  the  right 
half  are  equal  to  those  just  found  in  the  left. 

NOTE  2.  —  In  the  above  solution  we  have  called  forces  acting  up- 
wards and  to  the  right,  positive,  and  forces  acting  downwards  and  to 
the  left,  negative;  also  we  have  called  moments  tending  to  cause 
rotation  in  the  direction  of  the  hands  of  a  clock  from  left  to  right, 
positive,  and  those  in  the  opposite  direction,  negative.  The  opposite 
convention  would  do  as  well ;  we  have  only  to  introduce  opposite 
forces  and  opposite  moments  with  unlike  signs. 

Prob.  5.  A  roof  truss  like  Fig.  1  has  its  span  40  feet,  its 
rise  20  feet,  and  the  apex  loads  2000  Ibs. :  find  the  stresses 
in  ad,  ah,  he,  and  hd. 

Ans.  ad  =  +  1.5  tons,  ah  =  —  2.12  tons,  hc  =  —  1.41  tons, 
hd  =  -  0.71  tons. 

Prob.  6.  A  truss  like  Fig.  2  has  its  span  60  feet,  its  rise 
20  feet,  and  the  panel  loads  4000  Ibs. :  find  the  stresses  in 
ad,  ah,  he,  and  hd. 

Ans.  ad  =  +  4.5  tons,  ah  =  —  5.41  tons,  he  =  —  3.64  tons, 
hd  =  -  1.82  tons. 

Prob.  7.  A  truss  like  Fig.  3  has  its  span  80  feet,  its  rise 
20  feet,  and  the  panel  loads  10,000  Ibs. :  find  the  stresses 
in  all  the  members. 

Ans.  ak  =  +  15  tons,  kd  =  +  10  tons,  ah  =  —  16.77  tons, 
he  =  —14.5  tons,  hk=—  4.5  tons,  kc=+5  tons,  cd=00  tons. 


EOOF  TRUSSES.  11 

Art.  6.  Methods  of  Calculation.  —  Our  three  equa- 
tions of  equilibrium,  Art.  5,  furnish  us  with  two  methods 
of  calculation :  the  method  by  resolution  of  forces,  and  the 
method  of  moments.  The  principle  of  the  first  method  is 
embraced  in  the  first  two  equations  of  equilibrium ;  the 
principle  of  the  second  method  is  embraced  in  the  third 
equation  of  equilibrium. 

Either  of  these  methods  may  be  used  in  the  solution  of 
any  given  case ;  but  in  general  there  will  be  one,  the  em- 
ployment of  which  in  any  special  case  will  be  found  easier 
and  simpler  than  the  other.  Sometimes  a  combination  of 
both  methods  furnishes  a  readier  solution. 

REMARK.  —  A  section  may  be  passed  through  a  truss  in  any  direc- 
tion, separating  it  into  any  two  portions.  Thus,  in  Fig.  5,  a 
section  may  be  passed  around  h.  cutting  he,  hk,  and  ha.  Then  the 
internal  stresses  s±,  85.  sg,  and  the  apex  load  of  10,000  Ibs.  at  h, 
form  a  system  of  forces  in  equilibrium,  to  which  our  equations  are 
applicable. 

A  judicious  selection  of  directions  for  the  resolution  of  the  forces 
often  simplifies  the  determination  of  the  stresses.  Thus,  to  find 
85  in  Fig.  5,  if  we  resolve  the  forces  into  a  direction  perpendicular 
to  the  rafter  «c,  we  shall  obtain  an  equation  free  from  the  forces 
s3  and  s4;  whereas  if  the  directions  are  taken  at  random,  all  of 
the  forces  will  enter  the  equation.  This  principle  is  a  very  useful 
one. 

Thus,  we  have  at  once  in  Fig.  5,  calling  6  the  angle 
between  the  load  and  the  rafter, 

10000  sin  6  +  s5  =  0.      .-.  s5  =  - 10000  cos  cad  =  -  7895  Ibs. 

as  before. 

Similarly,  if  a  section  be  passed  around  d  in  Fig.  5,  cut- 
ting dk,  dc,  and  db,  and  the  vertical  components  be  taken, 
the  stress  in  dc  is  found  at  once  to  be  zero. 

In  solving  by  the  second  method,  the  center  of  moment  <•- 


12  ROOFS  AND  BRIDGES. 

may  be  taken  anywhere  in  the  plane  of  the  forces.  It  is 
often  more  convenient  to  write  three  moment  equations  for 
the  stresses  in  the  three  members  cut,  taking  a  new  center 
of  moments  each  time,  than  it  is  to  use  the  first  two  equa- 
tions of  equilibrium ;  and  if  the  center  of  moments  be  taken 
at  the  intersection  of  two  of  the  pieces  cut,  we  shall  have  at 
once  the  moment  of  the  stress  in  the  other  piece,  balanced 
by  the  sum  of  the  moments  of  the  external  forces,  since 
the  moments  of  the  stresses  in  the  other  two  cut  pieces  are 
zero. 

Thus,  in  solving  Prob.  4,  a  section  was  passed  cutting  s2, 
s6,  st.  To  find  s2  we  took  moments  about  c,  the  intersection 
of  s6  and  s^  which  gave  us  an  equation  of  moments  con- 
taining only  s2  and  known  terms.  To  find  s4  we  took 
moments  about  Jc,  the  intersection  of  s2  and  s6}  which  gave 
us  an  equation  of  moments  containing  only  s4  and  known 
terms.  Also,  to  find  s6  we  took  moments  about  a,  the  inter- 
section of  s2  and  s4,  which  gave  us  an  equation  of  moments 
containing  only  s6  and  known  terms. 

Therefore,  to  find  the  stress  in  any  member  by  the 
method  of  moments,  we  have  the  following  Rule: 

Conceive  at  any  point  of  this  member  a  section  to  be  passed 
completely  through  the  truss,  cutting  three  members.  To  find 
the  stress  in  this  member,  take  the  center  of  moments  at  the 
intersection  of  the  other  two.  Then  state  the  equation  of 
moments  betiveen  this  stress  and  the  external  forces  on  the 
left  of  the  section. 

Should  the  section  that  passes  completely  through  the 
truss  cut  more  than  three  pieces  whose  stresses  are  un- 
known, if  all  but  that  piece  in  which  the  stress  is  required 
meet  at  a  common  point,  the  center  of  moments  may  be 
taken  at  that  point. 


ROOF  TRUSSES.  13 

Should  the  section  cut  but  two  pieces,  the  center  of  mo- 
ments for  either  piece  may  be  taken  at  any  point  of  the 
other. 

Prob.  8.  With  the  dimensions  and  apex  loads  given  in 
Prob.  5  find  the  stress  in  cd  of  Fig.  1.  (See  Bern.) 

Ans.  +  1  ton, 

Prob.  9.  With  the  dimensions  and  apex  loads  given  in 
Prob.  6  find  the  stress  in  cd  of  Fig.  2.  Ans.  +  2.0  tons. 

Prob.  10.  In  a  truss  like  Fig.  4  the  span  is  80  feet,, 
the  rise  of  truss  is  20  feet,  the  rise  of  tie  rod  is  4  feet,  the 
panel  loads  are  each  4  tons :  find  the  stresses  in  all  the 
members. 

Ans.  ok  =  + 18.6  tons,  led  =  + 10  tons,  ah=—  20.6  tons, 
he  =  -  18.8  tons,  We  =  -  3.6  tons,  kc  =  +  9  tons,  cd  =  0 
(only  supports  part  of  the  tie  rod). 

Prob.  11.  In  Fig.  6  the  span  is  100  feet,  the  rise  of  truss 


is  25  feet,  the  apex  loads  are  3  tons:  find  the  stresses  in  all 
the  members. 

Ans.  ah=- 16.8,  hi  =  -  13.44,  Ic  =  -  10.08,  ale  =  + 15, 
Jem  =  -f- 15,  md  =  + 12,  hk  =  0  (only  supports  part  of  the 
tie),  Im  =  +  1.5,  cd  =  +  6,  hm  =  -  3.36,  Id  =  -  4.2. 

SUG. — The  stresses  in  the  lower  chords  and  in  all  the  verticals 
except  the  center  one  are  best  found  by  the  method  of  moments  ;  the 


14  ROOFS  AND  BRIDGES. 

stresses  in  the  upper  chords,  and  also  in  the  diagonals,  may  be  found 
by  the  method  by  resolution  of  forces,  although  the  stresses  in  the 
diagonals  are  easily  found  by  the  method  of  moments,  taking  the 
center  at  a.  To  find  the  stress  in  cd  it  is  best  to  pass  a  section  around 
d  and  take  vertical  components,  as  in  Probs.  8  and  9. 

Prob.  12.  A  truss  like  Fig.  6  has  80  feet  span,  20  feet 
rise,  distance  between  trusses  12  feet,  and  weight  of  roof  20 
Ibs.  per  scfuare  foot  of  roof  surface  (see  Prob.  1) :  find  the 
dead  load  stresses  in  all  the  members,  in  tons. 

Ana.  a/i  =  -11.5,  7tZ  =  -9.21,  fc  =  -6.9,  ok  =  +10.28  = 
Jem,  md=  +  8.22,  hk  =  0,  Im  =  + 1.03,  cd  =  +  4.11,  hm  = 
-2.3,  Id  =-2.88. 


Art.  7.    Lever  Arms  —  Indeterminate  Cases.  —  In 

determining  the  stresses  by  the  method  of  moments  the 
only  difficulty  lies  in  finding  the  lever  arms  of  the  various 
pieces.  These  can  always  be  found  by  the  use  of  geometry 
and  trigonometry.  Thus,  in  Fig.  6,  the  lever  arms  for  the 
lower  panels  are  evidently  the  perpendiculars  let  fall  upon 
these  panels  from  each  opposite  upper  apex.  The  lever 
arms  for  the  upper  panels  are  the  perpendiculars  drawn  to 
these  panels  from  each  opposite  lower  apex.  The  lever 
arm  for  each  brace  is  the  perpendicular  to  the  direction  of 
the  brace  drawn  from  the  left  end  a  of  the  truss,  where  the 
rafter  and  tie  intersect.  This  is  evident  from  our  rule  in 
Art.  6. 

Thus,  in  Fig.  6,  suppose  a  section  to  cut  hi,  hm,  and  km. 
By  our  rule  the  center  of  moments  for  km  is  h,  the  point  of 
intersection  of  the  other  two  pieces  hi  and  hm.  For  hi  the 
center  is  m,  the  intersection  of  hm  and  km.  For  hm  the 
center  is  a,  the  intersection  of  hi  and  km. 

For  trusses  in  which  the  members  have  various  inclina- 


HOOF  TRUSSES.  15 

tions,  all  different,  the  computation  of  the  lever  arms  is 
quite  tedious.  In  such  cases,  it  is  sometimes  advisable  to 
make  a  careful  drawing  of  the  truss,  and  then  measure  the 
lever  arms  by  scale.  Indeed,  this  method  can,  in  all  cases, 
be  used  as  a  check  upon  the  accuracy  of  the  results  obtained 
for  the  lever  arm  by  computation. 

If  the  section  dividing  the  truss  into  two  parts  cuts 
more  than  three  members,  the  stresses  in  which  are  un- 
known, the  problem  is  indeterminate,  because  there  are  more 
unknown  quantities  to  be  found  than  there  are  equations  of 
condition  between  them.  In  -such  cases  a  fourth  condi- 
tion is  sometimes  found  in  the  symmetry  of  the  truss  and 
loads. 

Thus,  if  we  pass  a  section  through  cl,  cd,  dl',  and  dm', 
Fig.  6,  it  cuts  more  than  three  pieces.  But  the  stress  in  dl' 
is  equal  to  that  in  dl,  by  reason  of  the  symmetry  of  the 
truss  and  loads.  Even  if  this  were  not  the  case,  it  could 
easily  be  found  by  working  toward  it  from  the  right  end. 
Then  there  remain  only  three  unknown  quantities,  whose 
values  can  be  determined  by  our  three  equations. 

The  section  may  cut  any  number  of  members,  so  long  as 
it  is  possible  to  find  independently  the  stresses  in  all  but 
three.  Any  truss  which  violates  this  rule  is  improperly 
framed,  and  has  unnecessary  or  superfluous  pieces. 

REMARK.  —  The  half  of  each  end  panel  load  carried  by  the  support 
does  not  affect  the  truss,  and  need  not  be  taken  into  account  in  find- 
ing either  loads  or  reactions.  Thus,  in  Fig.  5,  the  reaction  and  the 
half  panel  load  acting  at  the  support  are  equivalent  to  an  upward 
force  equal  to  their  difference.  This  upward  force  is  the  reaction 
found  by  omitting  the  two  apex  loads  at  the  supports,  and  is  known 
as  the  effective  or  working  reaction.  Thus,  instead  of  calling  20000 
the  reaction,  and  writing  the  equation 

20000  x  22$  -  5000  x  22$  -  si  x  17$  =  0, 


16 


ROOFS  AND  BRIDGES. 


as  is  done  in  the  solution  of  Prob.  4,  we  call  the  reaction  15000  and 
write  the  equation  as  follows  : 

15000  x  22|  -  si  x  17£  =  0, 

and  so  for  all  the  other  moment  equations. 

Prob.  13.   A  truss  like  Fig.  7  has  80  feet  span,  20  feet  rise, 
distance  between  trusses  12  feet,  and  weight  of  roof  as 


before  (see  Prob.  1)  :  find  the  dead  load  stresses  in  all  the 
members. 

Here  the  dead  apex  load 

2  x  20V52+102  =  3083  Ibs. 


=  1.54  tons,  say  1.5  tons. 

Effective  reaction  =  1.5  x  3.5  =  5.25  tons  (see  remark). 
To  find  fg  =  a/,  take  moments  around  6. 

5.25  x  10  -fg  x  5  =  0.     .\fg=+  10.5  tons. 
To  find  gh  take  moments  around  c. 
5.25  x  20  -  1.5  x  10  -  gh  X  10  =  0.     .-.  gh  =  +  9  tons. 
To  find  bg  pass  section  around  g  (see  rem.  of  Art.  6). 


—  9  =  0.     /.  bg=— 1.68  tons;  and  soon. 


ROOF  TRUSSES. 


17 


Ans.  ab=- 11.76,  be  =  -  10.08,  cd  =  -  8.4,  de  =  -  6.72, 
qf=  + 10.5,  fg  =  + 10.5,  grft  =  +  9,  fcfc  =  +  7.5,  6/=  0  (only 
supports  part  of  the  tie  rod),  eg  =  +  0.75,  dh  =  +  1.5, 
ek  =  +  4.5,  6gr  =  - 1.68,  eft  =  -  2.12,  dfc  =  -  2.7. 

Prob.  14.  In  Fig.  8  the  span  is  60  feet,  the  rise  of  truss 
is  15  feet,  the  rise  of  tie  rod  at  point  8  is  2  feet,  the  panel 


loads  are  1.6  tons,  the  struts  3-4,  5-6,  7-8  are  vertical, 
dividing  the  rafter  2-7  into  three  equal  parts:  find  the 
stresses  in  all  the  members. 

Ans.  Stress  in  2-3  =-9.75,  in  3-5  =-7.8,  in  5-7  = 
-5.85,  in  2-4= +8.73,  in  4-6  =  +  8.73,  in  6-8  =  +  7,  in 
3-4  =  0  (only  supports  part  of  tie  rod),  in  5-6  =  +  0.75, 
in  7-8  =  +  3.7,  in  3-6  =  -  1.85,  in  5-8  =  -  2.24. 

Prob.  15.  In  Fig.  9  the  span  is  90  feet,  the  rise  of  truss  is 
22.5  feet,  the  rise  of  tie  rod  is  3  feet,  the  struts,  3-4,  5-6, 


7-8,  and  9-10  are  vertical,  dividing  the  rafter  into  four 
equal  parts ;  the  apex  loads  are  2  tons :  find  the  stresses  in 
all  the  members. 


18  ROOFS  AND  BRIDGES. 

Ans.  Stress  in  2-3  =  — 18,  in  3-5  =  —  15.4,  in  5-7  =  — 
12.8,  in  7-9  =  -  10.2,  in  2^  =  +  16.1,  in  4-6  =  +  16.1,  in 
6-8  =+  13.8,  in  8-10  =  + 11.5,  in  3-4  =  0  (this  is  not 
necessary  to  the  stability  of  the  truss),  in  5-6  =  +  1.0,  in 
7-8  =+  2.0,  in  9-10  =  +  7.2,  in  3-6  =  -  2.46,  in  5-8  = 
-  2.96,  in  7-10  =  -  3.68. 

Prob.  16.  In  Fig.  10  the  span  is  120  feet,  the  rise  is  30 
feet,  the  struts  3-4,  5-6,  7-8  are  drawn  normal  to  the 


rafter,  dividing  it  into  four  equal  parts,  the  apex  loads  are 
2.5  tons :  find  the  stresses  in  all  the  members. 

Ans.  Stress  in  2-3  =  -  19.5,  3-5  =  -  18.45,  5-7  = 
-  17.33, 7-9  =  -  16.13,  2-4  =  +  17.5, 4-6  =  + 15.0, 6-10  = 
+10.0,  3-4  =  - 2.23,  5-6=  -4.45,  7-8=  -2.23,  4-5=  +2.5, 
5-8  =+  2.5,  6-8  =  +  5.0,  8-9  =  +  7.5,  9-10  =  0  (only  sup- 
ports part  of  the  tie  rod,  and  not  necessary  to  the  stability 
of  the  truss). 

It  will  be  observed  that  the  members  3-4  ttnd  7-8  are  symmetrical 
with  respect  to  their  loads,  and  therefore  their  stresses  are  equal,  and 
also  that  5-4  and  5-8  are  symmetrical,  and  hence  their  stresses  are 
equal. 

Prob.  17.  In  Fig.  11  the  span  is  90  feet,  the  rise  of  truss 
is  22.5  feet,  the  rise  of  tie  rod  is  3  feet,  the  struts  3-4,  5-6, 
7-8  are  drawn  normal  to  the  rafter,  dividing  it  into  four 
equal  parts ;  the  apex  loads  are  2  tons :  find  the  stresses  in 
all  the  members. 


ROOF  TRUSSES.  19 

Ans.  Stress  in  2-3  =  -20.34,  3-5=  - 19.44,  5-7=  -18.54, 
7-9  =  -17.64,  2-4  =  +  18.3,  4-6  =  +  15.66,  6-10  =  +  9.3, 
3^  =  -  1.78,  5-6  =  -  3.56,  7-8  =  -  1.78,  4-5  =  +  2.64, 
5_8  =  +  2.64,  6-8  =  +  6.84,  8-9  =  +  9.48,  9-10  =  0  (not 
necessary  to  stability  of  truss). 


Art.  8.  Snow  Load  Stresses.  —  The  snow  load  is 
estimated  per  square  foot  of  horizontal  projection  (Art.  3). 
If  the  main  rafters  of  the  truss  are  straight  —  as  in  all  of 
our  previous  problems  —  the  snow  load  is  uniformly  dis- 
tributed over  the  whole  roof,  and  the  apex  snow  loads  are 
all  equal  (see  Art.  4) ;  therefore  the  dead  load  and  snow 
load  stresses  are  proportional  to  the  corresponding  apex 
loads. 

Thus,  in  Prob.  13,  the  dead  panel  load  was  3083  Ibs.  and 
the  snow  panel  load  =  10  x  20  x  12  =  2400  Ibs.  Therefore 
if  we  multiply  each  dead  load  stress  by  !$$$(= -778),  we 
shall  have  the  corresponding  snow  load  stress. 

But  if  the  main  rafters  are  not  straight,  the  snow  load 
is  not  uniformly  distributed  over  the  whole  roof,  and  the 
apex  snow  loads  are  not  all  equal;  in  such  case,  the  snow 
load  stresses  have  to  be  determined  independently. 

Thus,  with  the  dimensions  given  in  Fig.  12,  for  9  feet 
between  trusses,  we  have  the  apex  snow  load  at  b  and  at  b' 
each  equal  to  one  half  the  panel  load  on  ab  plus  one  half 
the  panel  load  on  be 

=  6  x  20  x  9  =  1080  Ibs.  =  0.54  ton. 


20  ROOFS  AND  BRIDGES. 

The  apex  snow  load  at  c  equals  one  half  the  panel  load 
on  cb  plus  one  half  the  panel  load  on  cb' 

=  8  x  20  x  9  =  1440  Ibs.  =  0.72  ton. 

The  stresses  due  to  these  apex  snow  loads  may  now  be 
found  in  the  same  way  as  the  dead  load  stresses  were  found 
in  the  preceding  problems. 


Kig.  13 

It  is  possible  for  one  side  only  of  a  roof  to  be  loaded  with 
snow.  This  possibility  is  recognized  in  designing  roofs  of 
very  large  span,  such  as  the  roof  of  the  Jersey  City  train 
shed  of  the  Pennsylvania  E.R.  In  special  forms  of  trusses 
such  a  distribution  of  snow  load  may  produce  a  maximum 
stress  in  some  members. 

Problem,  With  the  dimensions  and  apex  snow  loads 
above  given,  find  the  snow  load  stresses  in  all  the  members 
of  Fig.  12. 

Ans.  a&=-1.44,  be  =  -1.3,  ad  =  +  0.85,  bd  =  +  0.58, 
dh  =  +  1.3,  dc  =  +  0.06. 

Art.  9.  Wind  Loads.  —  One  of  the  most  important 
questions  to  be  dealt  with  in  the  construction  of  roof-trusses 
is  the  pressure  of  the  wind;  for  it  is  evident  that  the 
stability  of  such  a  structure  depends  upon  its  power  of 
carrying  not  only  the  weight  of  the  truss,  roof,  snow,  etc., 
but  also  the  pressure  caused  by  the  wind.  In  the  case  of 
large  trusses,  it  will  often  be  found  that,  notwithstanding 
the  great  weight  of  the  structure,  the  stresses  produced  in 
some  of  the  members  by  a  gale  of  wind  are  almost  or  quite 


ROOF  TRUSSES.  21 

as  great  as  those  produced  in  the  same  pieces  by  the  dead 
load.  It  appears,  therefore,  that  in  designing  roofs,  especially 
iron  roofs  of  large  span,  a  correct  estimate  of  the  wind  loads 
is  quite  as  important  as  a  correct  estimate  of  the  dead  loads. 

And  yet,  the  subject  of  wind  forces  is  not  well  under- 
stood. It  is  hardly  possible  to  define,  with  any  precision, 
what  degree  of  violence  can  be  taken  to  represent  the 
greatest  wind  storm  that  has  to  be  provided  against.  In 
estimating  the  wind  pressure  and  the  resulting  stresses  in 
the  members  of  a  truss,  the  practice  of  engineers  has  varied 
greatly.  Until  recently,  it  has  been  considered  in  England 
sufficient  to  provide  for  a  wind  force  of  30  to  40  Ibs.  per 
square  foot  of  surface  normal  to  its  direction ;  while  in  this 
country  the  figures  have  been  taken  at  30  to  50  Ibs.  per 
square  foot. 

Taking  the  maximum  wind  pressure  against  a  surface 
normal  to  its  direction  as  50  Ibs.  per  square  foot,  we  shall 
probably  be  on  the  side  of  safety. 

The  following  table  gives  the  normal  wind  pressure  per 
square  foot  in  pounds  for  different  inclinations  of  the  roof 
equivalent  to  a  horizontal  wind  pressure  of  50  Ibs.  per 
square  foot,  calculated  by  Hutton's  formula : 

INCLIN.                   NOR.  PEES.  INCLIN.  NOR.  PEES. 

10°  12.1  33°  30' Q  span)  36.6 

15°                       18.0  35°  37.8 

20°                       22.6  40°  41.6 

21°  48'  Q  span)  25. 2  45°  43.0 

25°                       28.8  50°  47.6 

26°  34'  (J  span)  30.2  55°  49.5 

30°                       33.0  60°  50.0 

For  inclinations  greater  than  60°  the  normal  pressure  per 
square  foot  is  50  Ibs.  For  intermediate  inclinations  we  can 
find  the  pressures  by  interpolation. 


22  HOOFS  AND  BRIDGES. 

Art.  10.    Wind  Apex  Loads  and  Reactions.    Let 

Fig.  13  represent  a  roof  truss,  its  spaii  being  100  feet,  and 
its  rise  25  feet ;  let  the  distance  between  trusses  be  12  feet, 
and  suppose  the  wind  to  be  on  the  left  side.  Then  the 
inclination  of  the  rafter  ae  to  the  horizon  =  tan~H  =  26°34'. 
Hence,  from  our  table,  the  normal  wind  pressure  per  square 


Fig.  13 


foot  =  30.2  Ibs.  The  total  normal  wind  pressure  on  the 
side  of  the  roof  ae  is  therefore 

=  30.2  x  12  x  V502  +  25*  =  30.2  x  12  x  55.9  =  20258  Ibs., 

one  fourth  of  which,  or  5064.5  Ibs.,  is  the  pressure  on  each 
panel.  Hence  the  normal  wind  load  at  each  apex  6,  c,  d,  is 
5064.5  Ibs.,  or  say,  in  round  numbers,  5000  Ibs.,  or  2.5  tons, 
and  at  each  apex  a,  and  e,  it  is  2500  Ibs.,  or  1.25  tons 
(Art.  4). 

The  Reactions  caused  by  the  wind  pressure  are  inclined; 
the  horizontal  component  of  the  wind  tends  to  slide  the 
entire  truss  off  its  supports.  The  weight  of  the  truss  and 
the  roof  are  usually  sufficient  to  cause  friction  enough  to 
hold  it  in  place.  But  if  it  is  necessary,  the  truss  should  be 
fastened  at  its  ends  to  the  wall.  Eoof  trusses  of  short  span, 
and  especially  wooden  trusses,  have  generally  both  ends 
fixed  to  the  supporting  walls.  But  large  iron  trusses  have 
only  one  end  fixed,  while  the  other  end  is  free,  and  resting 
upon  friction  rollers,  so  that  it  may  move  horizontally, 


ROOF  TRUSSES.  23 

under  changes  of  temperature.  We  have  then  two  cases : 
the  first,  when  both  ends  are  fixed;  the  second,  when  one 
end  only  is  fixed,  and  the  other  end  is  free  to  move  upon 
rollers. 

CASE  I.  When  both  ends  are  fixed. — Let  Fig.  13  represent 
a  roof  truss  with  both  ends  fixed,  its  span  being  100  feet, 
its  rise  25  feet,  and  the  wind  apex  loads  1.25,  2.5,  2.5,  2.5, 
and  1.25  tons,  as  found  above.  In  this  case,  the  two  reac- 
tions El  and  R2  are  parallel  to  the  normal  wind  loads,  and 
may  easily  be  found,  as  follows : 

Let  0  be  the  angle  between  the  rafter  and  the  lower 
chord  ok,  and  take  moments  about  the  left  end  a.  We 
have  then 

R2  x  ah  -  (2.5  x  ab  +  2.5  x  ac  +  2.5  x  ad  + 1.25  x  ae)  =  0, 

or,  we  may  take  the  resultant  of  all  the  loads,  or  10  tons, 
acting  at  c, 

.-.  Rz  x  100  cos  e  -  10  x  27.95  =  0, 

.-.  R2  =  3.13  tons  (since  cos  0  =  ff). 

The  reaction  Rl  may  be  found  by  subtracting  R2  from, 
the  total  wind  load,  giving  us  Rl  =  6.87  tons.  Or,  we  may 
find  R!  by  taking  moments  about  the  right  end  7c.  Thus, 

R1X  ah  —  10  (ah  —  ac)  =  0, 
from  which  we  get  Rl  =  6.87  tons,  as  before. 

CASE  II.  When  one  end  is  fixed  and  the  other  is  free.  — 
(1)  Suppose  the  right  end  of  th«  truss  to  be  free,  and  the 
wind  blowing  on  the  fixed  side,  as  in  Fig.  14.  The  right  end 
of  the  truss  is  supposed  to  rest  upon  rollers,  the  support  at 
a  taking  all  the  horizontal  thrust  due  to  the  wind.  This 
being  the  case,  the  reaction  R.2  at  the  free  end  will  be  verti- 
cal, and  the  reaction  at  the  fixed  end  will  be  inclined. 


24  ROOFS  AND  BRIDGES. 

To  find  R2  take  moments  about  a ;  let  the  dimensions  and 
wind  loads  be  the  same  as  in  Fig.  13.     Thus, 

R2  x  100  - 10  x  27.95  =  0.        .-.  R2  =  2.8  tons. 


h      j       ft'     g'      f 

IB,  IB. 

Fig.  14 

Resolve  the  left  reaction  into  its  horizontal  and  vertical 
components,  H  and  R»  thus, 

2  hor.  comp.  =  0 
gives  10  sin 0-H=0.        .-.  #=4.46. 

2  ver.  comp.  =  0 

gives         R!  +  2.8  — 10  cos  0  =  0.        .-.1^  =  6.13. 
Or,  we  might  find  Ri  by  taking  moments  about  the  right 
support;  thus, 

.«!  x  100  - 10  (100  cos  6-28)  =  0.     .-.  ^  =  6.13,  as  before. 

(2)  Suppose  the  wind  to  blow  on  the  free  side  ek  of  the 
truss,  Fig.  14.  The  reaction  Rz  is  vertical,  as  before,  and 
the  reaction  at  the  fixed  end  a  may  be  resolved  into  its 
horizontal  and  vertical  components,  in  the  same  way  as 
above.  Thus,  we  find, 

H  =  4.46,     R1  =  2.8,     R2  =  6.13 ; 

that  is,  when  the  wind  changes  from  one  side  of  a  roof  to 
the  other,  the  horizontal  component  H  has  the  same  value 
as  before,  but  acts  in  the  opposite  direction,  and  the  re- 
actions RL  and  R2  interchange  their  values. 


ROOF  TRUSSES.  25 

Prob.  18.  A  truss  like  Fig.  14  has  its  span  40  feet,  rise 
10  feet,  and  the  total  normal  wind  load  on  the  fixed  side  3.2 
tons :  find  the  wind  load  reactions. 

Ans.  Rv  =  1.96,  R2  =  0.90,  H=  1.43  tons. 

Prob.  19.  A  truss  like  Fig.  3  has  its  span  50  feet,  its  rise 
12.5  feet,  and  the  distance  between  the  trusses  8  feet :  find 
the  reactions  when  both  ends  are  fixed. 

Ans.  Rl  =  2.32,  R2  =  1.06  tons. 

Prob.  20.  A  truss  like  Fig.  4,  with  one  end  free,  has  its 
span  80  feet,  its  rise  20  feet,  and  the  total  normal  wind  load 
on  the  fixed  side  5  tons :  find  the  wind  load  reactions. 

Ans.  Rl  =  3.07,  R2  =  1.4,  H=  2.23  tons. 

Prob.  21.  A  truss  like  Fig.  4,  with  one  end  free,  has  its 
span  60  feet,  rise  of  truss  12  feet,  rise  of  tie  rod  2  feet,  and 
the  total  normal  wind  load  on  the  fixed  side  4.5  tons :  find 
the  wind  load  reactions. 

Ans.  Ri  =  2.97,  Rz  =  1.21,  H=  1.67  tons. 

Prob.  22.  A  truss  like  Fig.  6,  with  one  end  free,  has  its 
span  90  feet,  its  rise  18  feet,  and  the  total  normal  wind  load 
on  the  fixed  side  6  tons :  find  the  wind  load  reactions. 

Ans.  Rv  =  3.96,  R2  =  1.62,  H  =  2.23  tons. 

Art.  11.  Wind  Stresses.  —  (1)  If  the  truss  have  both 
ends  fixed,  we  must  consider  the  wind  blowing  normally  to 
the  principal  rafters  on  one  side  only  (either  side  indiffer- 
ently) ;  and  as  the  wind  load  is  unsymmetrical  to  the  roof, 
the  wind  stresses  in  the  members  on  one  side  of  the  truss 
are  different  from  those  in  the  corresponding  members  on 
the  other  side,  and  hence  they  must  be  computed  for  every 
member  in  the  truss. 

(2)   In  trusses  with  one  end  fixed  and  the  other  free,  we 


26 


ROOFS  AND  BRIDGES. 


must  consider  the  wind  blowing  first  on  one  side  of  the 
truss,  and  then  on  the  other ;  and  the  stresses  produced  in 
the  two  cases  will  have  to  be  computed. 

Prdb.  23.  A  truss  like  Fig.  15,  with  both  ends  fixed,  has 
its  span  50  feet,  its  rise  12.5  feet,  and  the  wind  loads  and 
reactions  as  shown :  find  all  the  wind  stresses. 


We  find  ac  =  27.95  feet,  or,  in  round  numbers,  28  feet ; 
aJc  =  15.68  feet  =  Ice ;  hk  =  7  feet. 

Eepresenting  the  stresses  by  s1}  s2,  s3,  s4,  ss,  and  s6,  and 
applying  the  principles  of  Arts.  5  and  6,  we  have  for  the 
left  half  of  the  truss : 

2.62  x  14  -  *!  x  6.25  =  0,  .-.  8l  =  +  5.87  tons. 

2.62  x  28  -  3  x  14  -  s2  x  12.5  =0,  .-.  s2  =  +  2.51  tons. 

2.62xl4  +  s3x7       =0,  .-.  *s  =  —  5.24  tons  =  s4. 

3  +  s5              =0,  .-.  ss  =  -  3.00  tons. 

-  3  x  14  +  se  x  12.5  =  0,  .-.  s6  =  -f  3.36  tons. 

For  the  right  half  it  is  better  to  resolve  the  right  hand 
reaction  1.88  into  its  horizontal  and  vertical  components, 
thus, 

H  =1.88  sin  0  =  0.84,   and  V—  1.88  cos  0  =  1.68  tons, 

and  to  state  the  equation  of  each  piece  including  the  ex- 
ternal forces  on  the  right  of  the  section  rather  than  on  the 
left.  Thus, 


ROOF  TRUSSES.  27 

1.68  xl2.5-.84x6.25-*/x  6.25=0,  ...  8l'=  +2.52  tons. 

1.68  x25-.84x!2.5-s2'x  12.5=0,  .-.  *,'=  +2.52  tons. 

1.68xl5.68+s3'x7=0,  .-.  s3'=-3.76tons=s4'. 

ss'=0,  *6'=0. 

Prob.  24.  A  truss  like  Fig.  15,  with  one  end  free,  has  its 
span  40  feet,  its  rise  10  feet,  and  the  total  normal  wind  load 
on  the  fixed  side  3.2  tons :  find  all  the  wind  stresses. 

We  must  first  find  the  reactions  and  horizontal  compo- 
nent, as  in  Case  II.  Thus, 

R!  =  1.96,     E2  =  0.9,     H  =  1.43. 

Aw.  *!  =  +  3.6,  «2=  +  1.8,  s3=— 2.8,  s4=— 2.8,  s5=— 1.6, 
*8=+1.8,  a/ =+1.8,  s8' =  —2,  s/=-2,  s5'=0,  s6'=0. 

Sue. —It  will  often  be  best  to  state  the  equation,  using  the  ex- 
ternal forces  on  the  right  of  the  section.  Thus,  to  find  the  stress  in 
«2-  If  we  use  the  forces  on  the  right  of  the  section,  the  equation  is 

.9  x  20-s2x  10  =  0.         .-.  s2  =  1.8. 
But  if  we  use  the  forces  on  the  left  of  the  section,  the  equation  is 

1.96  x  20  +  1.43  x  10-32  x  11.18 -*a  x  10  =  0.        .-.  s2=  1.8. 
Of  course,  to  find  the  stresses  in  members  near  the  left  end,  not  so 
much  is  gained  by  using  the  forces  on  the  right  of  the  section. 

Prob.  25.  A  truss  like  Fig.  4,  with  one  end  free,  has  its 
span  60  feet,  its  rise  15  feet,  rise  of  tie  rod  3  feet,  and  the 
total  normal  wind  load  on  the  fixed  side  5  tons:  find  the 
wind  stresses  in  all  the  members. 

Am.  Stress  in  afc  =  +  8.7,  M=  +  3.5,  ah  =  —  7.78,  he 
=  -  7.78,  hk  =  -  2.5,  kc  =  +  5.45,  bk'  =  +  4.36,  bh'  =  -  4.8, 
ft'c  =  -4.8,  7t'fc'  =  0,  cfc'  =  +  1.0,  cd  =  0  (not  necessary  to 
stability  of  structure). 

Prob.  26.  A  truss  like  Fig.  14,  with  one  end  free,  has  its 
span  80  feet,  its  rise  20  feet,  and  the  total  normal  wind  load 


28  ROOFS  AND  BRIDGES. 

on  the  fixed  side  8  tons :  find  the  wind  stresses  in  all  the 
members. 

Ans.  Stress  in  af=  -f 11.16  =fg,  gli  =  +  8.92,  hj  =  +  6.68, 
bg  =  -  2.5,  ch  =  -  8.16,  dj  =  -4.02,  eg  =  +1.12,  dh  =  +  2.24, 
ej  =  +  3.35,  ab  =  -  9,  6c  =  —  7.5,  '  cd  =  —  6.0,  de  =  —  4.5, 
kf  =f'g'  =  gW  =  h'j  =  +  4.47,  6'flf'  =  c'A'  =  d'j  =  0,  6^  =  c'g' 
=  d'A'  =  0,  kb'  =  b'c'  =  c'd'  =  d'e  =  —  5.0  tons. 

Prob.  27.  A  truss  like  Fig.  16,  with  one  end  free,  has  its 
span  60  feet,  its  rise  12  feet,  the  struts  hk  and  h'k'  normal 


Fig.  1G 


L. 

JR. 


to  the  rafters  at  their  middle  points,  and  the  total  normal 
wind  load  on  the  fixed  side  ac  4  tons  :  find  the  wind 
stresses  in  all  the  members. 

Ans.  Stress  in  afc  =  +  5.4,  M  =  -(-2.7,  ah  =  —  4.6  =  he, 
hk  =  -2,  ck  =  +  2.7,  bk'  =  +  2.7  =  dk',  bh'  =  -  2.91  =  ch', 
h'k'  =  0  =  ck'. 

Prob.  28.  In  the  same  truss  Fig.  16,  with  the  same  dimen- 
sions as  given  in  Prob.  27,  let  the  same  normal  Avind  load  of 
4  tons  blow  on  the  free  side  be:  find  the  wind  stresses  in 
all  the  members. 

In  this  problem  the  values  of  the  reactions  are  the  same  as  those  in 
Prob.  27,  but  interchanged,  and  the  horizontal  component  has  the 
same  value,'but  acts  in  the  opposite  direction.  See  Case  II.  of  Art.  10  ; 
therefore,  here  RI  =  1.08,  R2  =  2.63,  H=  1.48. 


ROOF  TRUSSES.  29 

Ana.  Stress  in  6fc'  =  +  3.88,  eta' =+1.19,  6ft' =-4.58 
=  eft',  ft'fc'  =  -  2,  ck'  =  +  2.7,  a&  =  +  1.22,  dk  =  +  1.22,  aft 
=  -  2.91  =  eft,  lik  =  0  =  ck. 

Art.  12.  Complete  Calculation  of  a  Roof  Truss.  — 

By  comparison  of  the  values  of  the  stresses  in  Prob.  27  with 
those  in  Prob.  28,  we  see  that  the  stresses  are  quite  different, 
and  generally  greater  when  the  wind  blows  on  the  fixed  side 
of  the  roof  than  when  it  blows  on  the  free  side.  When  the 
wind  blows  on  the  fixed  side  it  tends  to  "  flatten  "  the  truss ; 
and  when  it  blows  on  the  free  side  it  tends  .to  "  shut  up  " 
the  truss,  or  "  double  it  up." 

In  the  complete  calculation  of  a  roof  truss,  we  must  find 
the  stresses  due  to  the  greatest  dead  load,  and  combine 
them  with  the  greatest  stresses  due  to  the  live  load,  so  as  to 
get  the  greatest  possible  tension  and  compression  in  each 
member.  If  the  dead  load  and  live  load  stresses  in  any 
piece  are  of  the  same  character,  both  compressive  or  both 
tensile,  we  must  add  them  to  obtain  the  greatest  stress  in 
the  piece.  But  if  these  stresses  are  of  opposite  characters, 
one  compression  and  the  other  tension,  their  difference  will 
be  the  resulting  stress  due  to  the  combination  of  live  and 
dead  loads,  and  if  the  live  load  stress  is  less  in  amount  than 
the  dead  load  stress,  it  will  only  tend,  when  the  wind  blows, 
to  relieve  the  stress  due  to  the  dead  load  by  that  amount, 
and  the  dead  load  stress  is  the  maximum  stress  in  the  mem- 
ber. But  if  the  live  load  stress  is  greater  than  the  dead 
load  stress  and  of  an  opposite  character,  it  will  cause  a 
reversal  of  stress  and  this  piece  will  then  need  to  be 
counterbraced. 

It  is  the  customary  American  practice  to  determine  the 
greatest  stresses  in  each  member  of  the  fixed  side  of  tho 
roof  truss  which  could  be  caused  by  the  wind  force  acting 


30  ROOFS   AND   BRIDGES. 

normally  to  the  truss  on  the  fixed  side  only,  and  then  to 
build  the  members  of  the  two  sides  of  the  truss  of  the  same 
size.  Since  the  stresses  caused  by  the  wind  blowing  on  the 
fixed  side  of  the  roof  are  at  least  as  great  as  those  caused 
by  the  wind  blowing  on  the  free  side,  this  arrangement 
gives  the  maximum  stresses,  and  is  on  the  safe  side  ;  and 
for  reasons  of  economical  manufacture  both  sides  of  the 
truss  are  constructed  alike. 

Prob.  29.  A  truss  like  Fig.  14,  with  one  end  free,  has  its 
span  80  feet,  its  rise  16  feet,  distance  apart  of  trusses  16 
feet,  rafter  divided  into  four  equal  parts,  struts  vertical, 
dead  load  of  roof  20  Ibs.  per  square  foot  of  roof  surface,  snow 
load  20  Ibs.  per  square  foot  of  horizontal  projection,  normal 
wind  load  on  fixed  side  by  table  of  Art.  9:  find  the  dead 
load,  snow  load,  wind  load,  and  maximum  stresses  in  all  the 
members. 

From  the  given  rise  and  span  we  have  the  length  of  one 
half  of  roof  ae=V402+162  =  8V29  =  43.08  feet. 

Weight  of  truss  =  ^  W  (Art.  2)  =  16       8°^  =  4266  Ibs. 


Dead  panel  load  =          +  48-08  ^  x  20  =  3977  Ibs. 

=  1.9885  tons,  or  say,  2  tons. 

Snow  load  per  panel  =  10  x  16  x  20  =  3200  Ibs.  =  1.6  tons. 
Inclin.  of  roof  =  tan-1  •  4  =  21°  48'. 
Nor.  pressure  of  wind  per  sq.  foot  (table  of  Art.  9)  =25.2  Ibs. 


ISTor.  wind  load  per  panel  =  25.2  x  X  16 

=  4342  Ibs.  =  2.171  tons,  or  say,  2.2  tons. 

From  these  data  the  dead  load,  snow  load,  and  wind  load 
stresses  (wind  on  the  fixed  side  only)  may  be  computed, 


ROOF  TRUSSES. 


31 


and  the  maximum  stresses  found,  for  all  the  members  of 
the  truss,  and  tabulated,  as  follows : 

STRESSES  IN  THE  LOWER  CHORD. 


MEMBERS. 

«/ 

fff 

gh 

A? 

Dead  load  stresses     .     . 
Snow  load  stresses     ... 
Wind  load  stresses     .     . 

+  17.50 
+  14.00 
+  14.80 

+  17.50 
+  14.00 
+  14.80 

+15.00 
+12.00 
+11.86 

+  12.50 
+  10.00 
+  8.88 

Maximum  stresses     .     . 

+46.30 

+  46.30 

+  38.86 

+31.38 

STRESSES  IN  THE  UPPER  CHORD. 


MEMBERS. 

db 

be 

cd 

de 

Dead  load  stresses     .     . 
Snow  load  stresses    .     . 
Wind  load  stresses    .     . 

-18.85 
-15.08 
-12.86 

-16.16 
-12.92 
-10.56 

-13.46 
-10.77 
-   8.25 

-10.77 
-    8.62 
-  5.94 

Maximum  stresses     .     . 

-46.79 

-39.64 

-32.48 

-25.33 

STRESSES  IN  THE  WEB  MEMBERS. 


MEMBERS. 

W 

dh      i        ej 

*ff 

ch 

dj 

Dead  load  stresses 
Snow  load  stresses 
Wind  load  stresses 

+  1.00 
+0.80 
+  1.19 

+  2.00 
+  1.60 
+  2.37 

+  6.00 
+  4.80 
+  3.56 

-2.68 
-2.14 
-3.19 

-3.20 

-2.56 
-3.78 

-  3.90 
-  3.12 
-  4.63 

Maximum  stresses 

+2.99 

+  5.97 

+  14.36 

-8.01 

-9.54 

-11.65 

(jbf  is  not  necessary  to  the  stability  of  the  structure.) 

Here  the  maximum  stress  of  each  kind  for  each  member 
in  the  windward  side  of  the  truss  is  found  by  adding  the 


32  ROOFS  AND  BRIDGES. 

dead  load,  snow  load,  and  wind  load  stresses,  giving  the 
greatest  total  tension  and  the  greatest  total  compression. 
Of  course,  the  wind  stresses  in  the  members  of  the  other 
half  of  the  truss  will  be  less  than  those  above  given  for  the 
members  of  the  half  on  the  windward  side,  and  therefore, 
the  maximum  stresses  in  these  members  will  be  less  than 
those  in  the  above  table,  since  the  dead  and  snow  load 
stresses  in  the  corresponding  members  of  the  fixed  and  free 
sides  of  the  truss  are  the  same. 

If  there  is  no  wind  blowing,  the  maximum  stresses  are 
found  by  adding  together  the  dead  load  and  snow  load 
stresses.  If  there  is  neither  wind  nor  snow  the  dead  load 
stresses  are  also  the  maximum  stresses.  If  the  wind  blows 
on  the  free  side  of  the  truss,  the  maximum  stresses  cannot 
exceed  those  found  above. 

Prob.  30.  A  truss  like  Fig.  9,  with  one  end  free,  has  its 
span  90  feet,  rise  of  truss  18  feet,  rise  of  tie  rod  3  feet, 
rafter  divided  into  four  equal  parts,  struts  vertical,  dead 
load  per  panel  2  tons,  snow  load  per  panel  1.5  tons,  normal 
wind  load  per  panel,  wind  on  fixed  side,  2.25  tons :  find  the 
dead  load,  snow  load,  wind  load,  and  maximum  stresses  in 
all  the  members  of  the  half  of  the  truss  on  the  windward 
side. 

Here  the  effective  reaction  for  dead  load 

=  2  x  3.5  =  7  tons. 

To  find  the  dead  load  stress  in  2-4  or  4-6,  take  moments 
around  the  point  3. 

.-.  dead  load  stress  in  2-4  =  7  X0^'25  =  21.00  tons. 
o.75 

Similarly  the  dead  load,  snow  load,  and  wind  load  stresses 
may  be  computed  for  all  the  members  of  the  truss. 


Ans. 


EOOF  TRUSSES. 
STRESSES  IN  THE  LOWER  CHORD. 


33 


MEMBERS. 

2-^ 

4-G 

c-s 

8-10 

Dead  load  stresses    .     . 
Snow  load  stresses    .     . 
Wind  load  stresses   .     . 

+  21.00 
+  15.75 
+  18.13 

+  21.00 
+  15.75 
+  18.13 

+  18.00 
+  13.50 
+  14.49 

+  15.00 
+  11.2o 
+  10.81 

Ma'ximum  stresses    .     . 

+54.88 

+  54.88 

+45.99 

+  37.09 

STRESSES  ix  THE  UPPER  CHORD. 


MEMBERS. 

2-3 

3-5 

5-7 

7-9 

Dead  load  stresses    .     . 
Snow  load  stresses    .     . 
Wind  load  stresses   .     . 

-22.60 
-16.95 
-16.29 

-19.36 
-14.52 
-13.28 

-16.12 
-12.09 
-10.26 

-12.88 
—  9.66 
-   7.74 

Maximum  stresses    .     . 

-55.84 

-47.16 

-38.47 

-30.28 

STRESSES  IN  THE  WEB  MEMBERS. 


MEMBERS. 

5-6 

7-3 

9-10 

3-6 

5-S 

7-10 

Dead  load  stresses 
Snow  load  stresses 
Wind  load  stresses 

+  1.00 
+0.75 
+  1.21 

+2.00 
+  1.50 
+2.43 

+  7.60 
+  5.70 
+   4.54 

-3.10 
-2.32 

-3.78 

-  3.50 
-  2.62 
-  4.23 

-  4.10 
-  3.08 
-  4.99 

Maximum  stresses 

+2.96 

+5.93 

+  17.84 

-9.20 

-10.35 

-12.17 

(ft/is  not  necessary  to  the  stability  of  the  truss.) 

Prob.  31.  A  truss  like  Fig.  10,  with  one  end  free,  has  its 
span  100  feet,  its  rise  20  feet,  the  rafter  divided  into  four 
equal  parts  by  struts  drawn  normal  to  it,  dead  load  per 
panel  2.5  tons,  snow  load  per  panel  1.5  tons,  normal  wind 
load  per  panel,  wind  on  fixed  side,  2.3  tons :  find  all  the 


ROOFS  AND  BRIDGES. 


stresses  in  all  the  members  of  the  half  of  the  truss  on  the 
windward  side. 


Ans. 


STRESSES  IN  THE  LOWER  CHORD. 


MEMBERS. 

2-4 

4-6 

6-10 

Dead  load  stresses     .     . 
Snow  load  stresses     .     . 
Wind  load  stresses     .     . 

+  21.83 
+  1:5.10 
+  15.64 

+  18.70 
+  11.22 
+  12.54 

+  12.50 
+   7.50 
+  6.21 

Maximum  stresses     .     . 

+  50.57 

+42.46 

+  26.21 

STRESSES  IN  THE  UPPER  CHORD. 


MEMBERS. 

£-3 

3-6 

5-7 

7-9 

Dead  load  stresses    .     . 
Snow  load  stresses    .     . 
Wind  load  stresses    .     . 

-23.50 
-14.10 
-13.57 

-22.57 
-13.55 
-13.57 

-21.65 
-12.99 
-13.57 

-20.73 
-12.44 
-13.57 

Maximum  stresses    .     . 

-51.17  • 

-49.69 

-48.21 

-46.74 

STRESSES  IN  THE  WEB  MEMBERS. 


MEMBERS. 

3-4  and 

5-6 

4-5  and 

5-8 

6-S 

8-9 

Dead  load  stresses 

-2.33 

-  4.65 

+3.10 

+  6.20 

+  9.30 

Snow  load  stresses 

-1.40 

-  2.79 

+  1.86 

+  3.72 

+  5.58 

Wind  load  stresses 

-2.30 

-  4.60 

+3.11 

+  6.21 

+  9.31 

Maximum  stresses 

-6.03 

-12.04 

+8.07 

+  16.13 

+24.19 

(9-10  is  not  necessary  to  the  stability  of  the  truss.) 

Prob.  32.  A  truss  like  Fig.  11,  with  one  end  free,  has  its 
span  100  feet,  the  rise  of  truss  20  feet,  the  rise  of  tie  rod 
2.5  feet,  the  rafter  divided  into  four  equal  parts  by  struts 
drawn  normal  to  it,  the  dead,  snow,  and  wind  loads  2.5 


EOOF  TRUSSES. 


35 


tons,  1.5  tons,  and  2.3  tons  respectively,  or  the  same  as  in 
Prob.  31 :  find  all  the  stresses  in  all  the  members  of  the 
half  of  the  truss  on  the  windward  side. 


Ans. 


STRESSES  IN  THE  LOWER  CHORD. 


MEMBERS. 

2-4 

4-6 

6-10 

Dead  load  stresses     .     . 
Snow  load  stresses    .     . 
Wind  load  stresses    .     . 

+28.55 
+  17.13 
+  19.92 

+24.50 
+  14.70 
+  17.30 

+  14.35 
+   8.61 
+   7.00 

Maximum  stresses     .     . 

+  65.60 

+  56.50 

+29.96 

STRESSES  IN  THE  UPPER  CHORD. 


MEMBERS. 

2-3 

8-5 

5-7 

7-9 

Dead  load  stresses    .     . 
Snow  load  stresses   .     . 
Wind  load  stresses    .     . 

-30.60 
-18.36 
-18.12 

-19.68 
-]7.81 
-18.12 

-28.75 
-17.25 
-18.12 

-27.83 
-16.70 
-18.12 

Maximum  stresses    .     . 

-67.08 

-55.61 

-64.12 

-62.65 

STRESSES  IN  THE  WEB  MEMBERS. 


MEMBERS. 

3-4  and 

7-8 

5-6 

4-5  and 
5-8 

6-8 

8-9 

Dead  load  stresses 

-2.33 

-  4.65 

+  4.00 

+  10.65 

+  14.65 

Snow  load  stresses 

-1.40 

-  2.79 

+  2.40 

+  6.39 

+  8.79 

Wind  load  stresses 

-2.30 

-  4.60 

+  4.02 

+  9.15 

+  12.10 

Maximum  stresses 

-6.03 

-12.04 

+  10.42 

+26.19 

+  36.60 

(9-10  is  not  necessary  to  the  stability  of  the  truss. ) 

Prob.  33.  A  truss  like  Fig.  11,  with  one  end  free,  has  its 
span  120  feet,  rise  of  truss  20  feet,  rise  of  tie  rod  3  feet, 
rafter  divided  into  four  equal  parts  by  struts  drawn  normal 


36 


EOOFS  AND  BRIDGES. 


to  it,  distance  between  trusses  20  feet,  dead  load  of  roof 
15  Ibs.  per  square  foot  of  roof  surface,  snow  load  20  Ibs. 
per  square  foot  of  horizontal  projection,  normal  wind  load 
on  fixed  side  by  table  of  Art.  9 :  find  all  the  stresses  in  all 
the  members  in  the  half  of  the  truss  on  the  fixed  side. 

Ans.  Dead  load  per  panel  =  3.1  tons ;   snow  load  per 
panel  =  3  tons ;  wind  load  per  panel  =  3.3  tons. 

STRESSES  IN  THE  LOWER  CHORD. 


MEMBERS. 

2-4 

4-6 

6-10 

Dead  load  stresses     .     . 
Snow  load  stresses    .    •. 
Wind  load  stresses    .     . 

+  45.26 
+  43.80 
+  37.35 

+  38.91 
+  37.65 
+  29.92 

+  22.17 
+21.45 
+  12.66 

Maximum  stresses     .     . 

+  126.41 

+  106.48 

+56.18 

STRESSES  IN  THE  UPPER  CHORD. 


MEMBERS. 

2-3 

3-5 

5-7 

7-9 

Dead  load  stresses    .     . 
Snow  load  stresses   .     . 
Wind  load  stresses   .     . 

-  47.52 
-  45.99 
-  35.31 

-  46.56 
-  45.06 
-  36.31 

-  45.60 
-  44.13 
-  36.31 

-  44.64 
-  43.20 
-  35.31 

Maximum  stresses    .     . 

-128.82 

-126.93 

-125.04 

-123.15 

STRESSES  IN  THE  WEB  MEMBERS. 


MEMBERS. 

3^  and 

7-8 

5-6 

4-5  and 
5-8 

6-8 

8-9 

Dead  load  stresses 

-2.95 

-  6.89 

+    6.36 

+  17.3S 

+  23.72 

Snow  load  stresses 

-2.85 

-  5.70 

+  6.15 

+  16.80 

+22.95 

Wind  load  stresses 

-3.36 

-  6.70 

+  7.44 

+  17.76 

+25.19 

Maximum  stresses 

-9.15 

-18.29 

+  19.95 

+  51.92 

+  71.86 

ROOF  TEUSSES.  37 

When  the  members  of  a  truss  have  many  different  in- 
clinations, as,  for  example,  in  the  Crescent  Truss  and  Arch 
Truss,  the  calculation  of  the  stresses  by  the  method  of 
moments  and  the  method  by  resolution  of  forces  (Art.  6), 
becomes  very  tedious,  owing  to  the  difficulty  in  finding  the 
lever  arms,  or  the  sines  and  cosines.  In  practice  the 
graphic  method  is  often  used  for  determining  the  stresses 
in  such  trusses ;  but  if  the  analytic  method  is  preferred  to 
the  graphic,  the  truss  should  be  drawn  to  a  large  scale,  and 
all  the  lever  arms  and  sines  and  cosines  be  scaled  from  the 
diagram. 

The  present  chapter  is  but  a  brief  treatise  on  roof  trusses. 
The  student  who  desires  to  continue  the  subject  further  is 
referred  to  more  extended  works,  such  as  "Strains  in 
Framed  Structures,"  by  Du  Bois,  and  "Theory  and  Prac- 
tice of  Modern  Framed  Structures,"  by  Johnson,  Bryan, 
and  Turneaure. 


CHAPTER   II. 

BRIDGE  TRUSSES  WITH  UNIFORM   LOADS. 

Art.  13.  Definitions.  —  A  Bridge  is  a  structure  for 
carrying  moving  loads  over  a  body  of  water,  or  over  a 
depression  in  the  earth.  For  railroad  and  highway  accom- 
modations, it  connects  two  roads  so  as  to  form  a  continuous 
road.  A  framed  bridge,  or  trussed  bridge  is  composed  of  two 
or  more  trusses,  which  lie  in  vertical  planes,  parallel  to  the 
line  of  the  road.  A  bridge  truss,  like  a  roof  truss,  consists 
of  the  upper  and  lower  chords  and  the  web  members.  The 
upper  chord  of  a  simple*  bridge  truss  is  in  compression,  and 
the  lower  is  in  tension,  while  the  web  members  are  some  in 
compression  and  some  in  tension.  The  trusses  are  united 
by  lateral  bracing,  attached  to  the  panel  points  of  either  the 
upper  or  lower  chords,  or  both.  The  object  of  this  lateral 
bracing  is  to  support  the  trusses  sideways,  and  stiffen  the 
structure  against  the  action  of  the  wind. 

The  Floor,  or  Floor  System,  of  a  bridge  consists  of 
the  floor  beams,  stringers,  and  flooring.  The  floor  beams  run 
at  right  angles  with  the  chords,  and  are  connected  to  them 
at  the  panel  points.  The  stringers  frame  into  or  rest  upon 
the  floor  beams,  and  are  parallel  to  the  chords,  and  support 
the  flooring  of  a  highway  bridge,  or  the  cross  ties  and  rails 
of  a  railroad  bridge. 

*This  is  true  only  for  simple  trusses,  and  not  for  cantilever  trusses, 
continuous  trusses,  or  trusses  of  draw  spans. 


BRIDGE   TRUSSES. 


39 


A  Through  Bridge  is  one  in  which  the  roadway  is  sup- 
ported by  the  bottom  chords,  with  lateral  bracing  overhead 
between  the  top  chords. 

A  Deck-bridge  is  one  in  which  the  roadway  is  sup- 
ported by  the  top  chords ;  the  trusses  are  usually  placed 
nearer  to  each  other  than  on  through  bridges,  the  roadway 
extending  over  them. 

When  a  through  bridge  is  not  of  sufficient  height  to 
allow  of  the  upper  lateral  bracing,  the  trusses  are  called 
Pony  Trusses.  Such  trusses  are  necessarily  short;  they  are 
'<  stayed  "  by  bracing  connected  with  the  floor  system. 


Art.  14.  Different  Forms  of  Trusses.  —  The  King- 
post Truss  is  a  term  applied  to  a  truss  in  which  there  is  a/ 
central  vertical  tie  and  two  braces  resting  against  it,  as  in 
Fig.  17.  It  was  formerly  built  for  country  highway  bridges 


d 
Fig.  IT 


<«srA 
&. 


of  short  span.  The  term  "king-post"  is  an  old  one,  and 
came  into  use  when  the  central  piece  was  made  of  wood, 
and  resembled  a, post,  although  its  office  was  that  of  a  tie; 
this  tie  is  now  usually  a  wrought  iron  rod.  The  load  at  the 
center  d  is  carried  by  the  tie  up  to  the  apex  c,  and  then  by 
the  two  inclined  braces  ca  and  cb  down  to  the  abutments,  a 
and  6. 

The  Queen-post  Truss  is  a  truss  in  which  there  are 
two  vertical  ties  against  which  rest  two  braces,  as  in  Fig.  18. 


40 


ROOFS  AND  BRIDGES. 


This  is  also  an  old  term  and  has  become  familiar  with  long 
use.     It  is  sometimes  called  a  trapezoidal  truss.     The  panel 


loads  at  h  and  k  are  carried  up  the  ties  to  c  and  d,  and  then 
by  the  two  inclined  struts  ca  and  db  down  to  the  abutments, 
a  and  b. 

The  Howe  Truss,  Fig.  19,  has  its  vertical  members  in 
tension  and  the  inclined  ones  in  compression ;  the  diagonal 
counter  struts  are  in  broken  lines.  The  chords  and  diagonal 


Kig.  10 

web  members  are  of  wood,  and  the  vertical  ties  of  iron. 
This  truss  was  patented  in  the  United  States  in  1840  by 
William  Howe.  It  has  proved  the  most  useful  style  of 
bridge  truss  ever  devised  for  use  in  a  new  and  timbered 
country.  It  is  still  very  largely  used  where  timber  is 
cheap,  for  both  highway  and  railway  bridges. 

The  Pratt  Truss,  Fig.  20,  has  its  vertical  members  in 
compression   and  the  inclined   ones   in   tension.     All   the 


Fig.  SO 

members  of  this  truss  are  of  iron  or  steel ;  though  it  was 
formerly  built  all  of  wood,  except  the  diagonal  ties,  which 


BRIDGE  TRUSSES.  41 

were  of  wrought  iron.  The  Pratt  truss  is  a  favorite  type, 
and  is  used  more  than  any  other  kind.  Figure  20  shows  a 
deck-bridge.  The  deck  or  through  Pratt  truss  is  the  stand- 
ard form  of  truss  for  both  highway  and  railway  bridges  of 
moderate  spans,  though  it  is  not  generally  used  for  railway 
bridges  in  which  the  span  is  much  less  than  100  feet. 

The  Warren  Truss,  or  Warren  Girder,  has  all  its 
web  members  inclined  at  equal  angles,  some  of  them  being 
in  tension  and  some  in  compression.  This  truss  is  an 
example  of  the  pure  triangular  type.  Its  web  members 
consist  always  of  equilateral  triangles.  When  the  triangles 
are  not  equilateral,  the  truss  is  simply  a  "  triangular  truss." 


•vwwvw* 


Figure  21  shows  a  Warren  truss  as  a  deck  truss.  The  War- 
ren truss  is  generally  built  all  of  iron  or  steel,  and  is  used 
for  comparatively  short  spans;  it  is  of  more  frequent 
occurrence  in  England  than  in  this  country. 

The  Double  Triangular,  or  Double  Warren  Truss 
(or  Girder),  Fig.  22,  has  each'  panel  braced  with  two 
diagonals  intersecting  each  other,  and  forming  a  single 
lattice. 


XXXXXXXXXXXXXXXX 


Fig. 


Other  forms  of  trusses  will  be  explained  as  we  proceed. 
All  these  forms  of  trusses,  and  bridge  trusses  generally, 
may  be  arranged  so  as  to  be  used  either  for  deck-bridges 
or  through  bridges. 


42  ROOFS  AND  BRIDGES. 

Art.  15.  The  Dead  Load,  or  permanent  load,  consists 
of  the  entire  weight  of  the  bridge.  It  includes  the  weight 
of  the  trusses,  the  lateral  bracing,  and  the  floor  system.  In 
highway  bridges,  the  floor  system  consists  of  the  floor  beams, 
which  are  supported  by  the  chords  at  the  panel  points,  the 
stringers,  which  are  supported  by  the  floor  beams,  and  the 
planks,  which  are  supported  by  the  stringers.  In  railway 
bridges  the  stringers  support  the  cross  ties,  rails,  guard- 
rails, spikes,  etc.  The  dead  load  depends  upon  the  length 
of  the  bridge,  its  width,  its  style,  and  upon  the  live  loads  it 
is  intended  to  carry.  For  highway  bridges  with  plank 
floors  the  total  dead  load  per  foot  may  be  found  approxi- 
mately by  the  following  formula : 

to  =  150  +  cbl  +  4  bt, 

where  w  =  weight  in  pounds  per  linear  foot  of  bridge. 
I  =  length  of  bridge  in  feet, 
b  =  width  of  roadway  in  feet, 
t  =  thickness  of  planking  in  inches, 
c  =  \  for  heavy  city  bridges, 
=  \  for  ordinary  city  or  suburban  bridges, 
=  \  for  light  country  bridges. 

For  single  track  railroad  bridges  the  dead  load  per  linear 
foot  is  given  very  closely  by  the  following  formulae  (taken 
from  "Modern  Framed  Structures,"  by  Johnson,  Bryan, 
and  Turneaure,  p.  44). 

For  deck-plate  girders, 

?(,  =  9Z  +  520  ........     (1) 

For  lattice  girders, 

w  =  7 1  +  600 (2) 


BRIDGE   TRUSSES.  43 

v\ 

For  through  pin-connected  bridges, 

w  =  5 1  +  750 (3) 

For  Howe  trusses,     w  =  6.5l  +  675 (4) 

where  I  is  the  span  in  feet,  and  w  the  dead  load  in  pounds 
per  linear  foot.  These  four  formulas  give  dead  loads  of 
iron  railway  bridges,  including  an  allowance  of  400  Ibs.  per 
foot  for  the  weight  of  track  material,  for  bridges  designed 
to  carry  100-ton  locomotives.  For  lighter  locomotives  the 
dead  load  would  be  less,  and  for  heavier  locomotives  it 
would  be  more.  For  double  track  bridges  add  90  per  cent 
to  the  above  values ;  *  for  the  load  on  each  truss  take  one 
half  the  above  values. 

Prob.  34.  What  is  the  weight  of  a  Warren  truss  bridge 
105  feet  long  ? 

Here  we  may  find  the  dead  load  from  formula  (2),  as 
follows : 

Dead  load  per  linear  foot 

=  w  =  7  I  +  600  =  1335  Ibs. 

.-.  weight  of  bridge  =  1335  x  105  =  140,175  Ibs.  and 
weight  of  bridge  to  be  carried  by  one  truss 

=  70087.5  Ibs. 

Prob.  35.   What  is  the  weight  of  a  through  pin-connected 
bridge  of  100  feet  span  ? 
Ans.  125,000  Ibs. 

Prob.  36.   What  is  the  weight  of  a  Howe  truss  bridge  of 
144  feet  span  ? 
Ans.  231,984  Ibs. ;  or  115,992  Ibs.  per  truss. 

*For  double  track  bridges  the  weight  of  the  metal  work  is  about  90% 
greater  than  for  a  single  track ;  but  the  weight  of  the  track  material  is 
just  double  that  for  a  single  track  bridge. 


44 


ROOFS  AND  BRIDGES. 


Art.  16.  The  Live  Load,  or  moving  load,  is  that  which 
moves  over  the  bridge,  and  consists  of  wagons  and  foot  pas- 
sengers on  highway  bridges,  and  trains  on  railway  bridges. 

For  highway  bridges,  the  live  load  is  usually  taken  as  a 
uniform  load  of  from  50  to  100  Ibs.  per  square  foot  of  road- 
way, or  the  heaviest  concentrated  load,  due  to  a  road-roller, 
which  is  likely  to  come  upon  the  bridge.  The  trusses  of 
highway  bridges  are  usually  found  to  receive  the  greatest 
stresses  from  a  densely  packed  crowd  of  people,  while  the 
floor  system  usually  receives  the  greatest  stresses  from  the 
concentrated  load.  The  following  are  the  live  loads  speci- 
fied by  Waddell,  per  square  foot  of  floor : 


Span  of  Bridge. 

City  and  Suburban  Bridges. 

Country  Bridges. 

0  to    50  feet. 

100  Ibs. 

90  Ibs. 

50  to  150  feet. 

90  Ibs. 

80  Ibs. 

150  to  200  feet. 

80  Ibs. 

70  Ibs. 

200  to  300  feet. 

70  Ibs. 

60  Ibs. 

300  to  400  feet. 

60  Ibs. 

50  Ibs. 

The  live  load  per  panel  of  a  highway  bridge  may  then  be 
found  by  multiplying  the  width  of  roadway  including  the 
sidewalks  by  the  product  of  the  load  per  square  foot  with 
the  panel  length. 

For  railway  bridges,  the  live  load  is  often  specified  as  the 
weight  of  two  of  the  heaviest  locomotives  which  we  think 
will  ever  pass  over  the  bridge,  and  followed  by  a  uniform 
load  due  to  the  heaviest  possible  train. 

In  English,  and  sometimes  in  American  practice,  it  is 
considered  sufficiently  accurate  to  use  the  corresponding 
uniformly  distributed  load  which  causes  the  same  stresses 
in  the  chords  as  those  which  result  from  the  above  con- 
centrated locomotive  loads. 


BRIDGE   TRUSSES.  45 

The  following  equivalent  uniformly  distributed  live  loads 
per  linear  foot  of  track,  produce  approximately  the  same 
stresses  as  the  above  concentrated  loads. 

Span,  10,  20,  30,  40,  50,  100,  200,  300  feet. 

Load,  10,000,  6600,  5500,  4900,  4600,  4000,  3700,  3500  Ibs. 

This  loading  is  very  nearly  that  used  by  the  Pennsyl- 
vania Railroad.  The  Erie  Railroad  uses  a  live  load  one  fifth 
greater;  and  the  Lehigh  Valley  Eailroad  a  live  load  one 
third  greater. 

The  live  load  is  taken  greater  for  short  spans  than  for 
long  ones,  because  if  the  span  is  short  one  or  two  locomo- 
tives may  cover  the  whole  bridge,  while  if  the  span  is  long, 
the  whole  bridge  would  not  often  be  loaded  with  more  than 
a  train  drawn  by  one  or  two  locomotives.  The  calculations 
of  the  stresses  are  made  in  precisely  the  same  way  for 
railway  as  for  highway  bridges,  so  long  as  the  live  load  is 
uniform. 

In  the  use  of  equivalent  uniform  loads,  the  following  precautions 
are  to  be  observed :  In  obtaining  the  stresses  in  the  chords  and  the 
main  web  members,  the  equivalent  load  corresponding  to  the  length 
of  the  truss  is  to  be  used.  But  in  such  members  as  receive  a  maxi- 
mum stress  from  a  single  panel  load,  another  equivalent  load  must  be 
used.  Thus,  in  the  "hip  verticals"  of  a  Pratt  truss,  Fig.  30,  or  in 
the  "vertical  suspenders"  of  a  Warren  truss,  Fig.  35,  the  maximum 
stress  is  obtained  by  using  the  equivalent  load  corresponding  to  a 
span  of  two  panel  lengths. 

Prob.  37.  A  bridge  for  a  city  has  its  roadway  20  feet 
wide  in  the  clear,  and  also  two  sidewalks,  each  6  feet 
wide  in  the  clear.  The  span  is  200  feet  and  there  are  10 
panels.  Find  the  live  panel  load  per  truss. 

Am.  12.8  tons. 


\ 
46  ROOFS  AND  BRIDGES. 

Prob.  38.  A  country  bridge,  50  feet  long,  has  its  roadway 
16  feet  wide  in  the  clear,  and  a  sidewalk  8  feet  wide  in  the 
clear.  There  are  5  panels.  Find  the  live  panel  load  per 
truss. 

Ans.   5.4  tons. 

Art.  17.  Shear — Shearing  Stress. — Let  Fig.  23  repre- 
sent a  beam  fixed  horizontally  at  one  end  and  sustaining  a 
load  P  at  the  other  end.     Imagine 
the  beam  divided  into  vertical  slices 
or  transverse  sections  of  small  thick- 
ness.    The  weight  P  tends  to  sepa- 
rate or  shear  the  section  or  slice  on 
which  it  immediately  rests  from  the 
Fig.  as  p     adjoining   one.       The    lateral    con- 

nection of  the  sections  prevents  this 

separation,  and  the  second  section  or  slice  is  drawn  by  a 
vertical  force  equal  to  the  weight  P  which  tends  to  slide  or 
shear  it  from  the  third  section,  and  so  on.  Thus,  a  vertical 
force  equal  to  the  weight  P  is  transmitted  from  section  to 
section  throughout  the  length  of  the  beam  to  the  point  of 
support  This  vertical  force  is  called  the  "  shearing  force," 
or  "shear" ;  and  the  equal  and  opposite  internal  force  or 
stress  in  the  section  that  balances  it  is  called  the  "  shearing 
stress." 

The  shear  then  at  any  section  is  that  force  which  tends  to 
make  that  section  slide  upon  the  one  immediately  following. 

The  vertical  shear  at  any  section  is  the  total  vertical  force 
at  that  section,  and  it  is  equal  to  the  sum  of  the  vertical  com- 
ponents of  all  the  external  forces  acting  upon  the  beam  on 
either  side  of  the  section. 

Thus,  let  Fig.  24  represent  a  beam  I  feet  long,  resting 
horizontally  on  supports  at  its  extremities;  and  let  the 


BRIDGE   TRUSSES.  47 

beam  have  a  uniform  load  of  w  Ibs.  per  foot.  Consider  a 
section  ab  at  any  distance  x  from  the  left  end.  Then,  the 
reaction  at  each  support  is  %  wl ;  the  shear  in  the  section 
ab  is  the  weight  that  is  between  that  section  and  the  center 


Fig.  24 

of  the  beam,  or  (^  I  —  x)w,  since  this  is  the  total  vertical 
force  at  ab,  which  is  the  shear  by  definition ;  but  this  is  the 
same  as  the  algebraic  sum  of  the  vertical  components  on 
the  left  side  of  ab. 

For  the  algebraic  sum  of  the  vertical  components  on  the 
left  of  the  section  ab  =  \  Iw  —  xw,  or  (i  I  —  x)w. 

Art.  18.  Web  Stresses  due  to  Dead  Loads  —  Hori- 
zontal Chords.  —  Let  ABCD  be  a  truss  with  horizontal 


chords*  AB,  CD,  and  let  a  section  mn  be  drawn  in  any 
panel  be  cutting  the  diagonal  bh  and  the  two  chords,  and 
let  9  be  the  angle  which  the  diagonal  makes  with  the  ver- 
tical. Then  (Art.  5)  the  stresses  in  these  three  members 
must  be  in  equilibrium  with  the  external  forces  on  the  left 
of  the  section;  and  therefore  the  algebraic  sum  of  the 

*  Called  also  flanges. 


48  ROOFS  AND  BRIDGES. 

vertical  components  of  these  forces  equals  zero.     Hence, 
calling  S  the  stress  in  the  diagonal  bh,  we  have 

R-Pl-Pt  +  Scos6  =  0 (1) 

But  R  —  Pl  —  P2  is  the  vertical  shear  for  the  given  section 
mn  (Art.  17). 

Hence  (1)  becomes 

Shear  +  S  cos  0=0; 

.-.  S  =  —  shear  sec  0 (2) 

Therefore,  for  horizontal  chords  and  vertical  loads,  the 
stress  in  any  web  member  is  equal  to  the  vertical  shear  multi- 
plied by  the  secant  of  the  angle  which  the  member  makes  with 
the  vertical. 

Since  the  shear  in  the  section  cutting  he  is  the  same  as 
that  in  the  section  mn,  no  load  being  at  h,  therefore  the 
stress  in  ch  is  equal  to  the  shear  in  ch,  that  is,  equal  to  the 
shear  in  the  panel  be.  Since  there  are  no  loads  between 
the  joints  6  and  c,  the  shear  is  constant  throughout  the 
panel  be;  and  we  usually  speak  of  it  as  the  shear  in  the 
panel  be. 

It  will  be  observed  from  (2)  that,  for  the  diagonal  bh, 
the  stress  is  negative,  or  compre*ssive,  provided  that  the 
shear  is  positive;  but  for  the  dead  load  the  shear  is  always 
positive  in  sections  left  of  the  middle  of  the  truss.  Hence 
the  stress  in  bh  is  compressive,  and  so  for  all  the  other 
diagonals  in  the  left  half  of  the  truss,  since  they  are  all 
inclined  in  the  same  direction,  that  is,  downwards  toward 
the  left.  Conversely,  for  members  inclining  .downwards 
toward  the  right,  in  the  left  half  of  the  truss,  the  stresses 
are  positive  or  tensile. 

Prob.  39.  A  through  Howe  truss,  like  Fig.  25,  has  8 
panels,  each  15  feet  long,  and  20  feet  deep:  find  all  the 


BRIDGE   TRUSSES.  49 

web  stresses  due  to  a  dead  load  of  450  Ibs.  per  linear  foot 
per  truss. 

Panel  load  =  45°*15  =      =  3.38  tons. 


Keaotion    =      x    =        =  11.8  tons, 

which  is  also  the  shear  in  panel  Aa.  The  shear  in  each  of 
the  other  panels  is  found  by  subtracting  from  the  reaction 
the  loads  on  the  left  of  the  panel. 


20 

The  secant  for  the  verticals  =  1. 
We  have  then  the  following  stresses  for  the  diagonals: 

Stress  in  AC  =  -  11.8  x  1.25  =  -  14.8  tons. 
Stress  in  ae  =  -  (11.8  -  3.38)1.25  =  -  10.5  tons. 
Stress  in  bh  =  -  (11.8  -  6.76)1.25  =  -  6.3  tons. 
Stress  in  ck  =  -  (11.8  -  10.14)1.25  =  -  2.1  tons. 

For  the  verticals : 

Stress  in  aC=  + 11.8  tons.     Stress  in  be  =  +  8.4  tons. 
Stress  in  ch  =  +    5.1  tons.     Stress  in  dk  =  +  3.4  tons. 

This  last  value  is  found  by  passing  a  section  around  d  cut- 
ting dc,  dk,  dc'  (see  Rem.  of  Art.  5),  and  taking  vertical 
components. 

Prob.  40.   A  deck  Pratt  truss,  like  Fig.  20,  has  12  panels, 
each  6  feet  long,  and  6  feet  deep:  find  all  the  web  stresses 
due  to  a  dead  load  of  500  Ibs.  per  linear  foot  per  truss. 
Ans.   Stresses  in  verticals 

=  _8.25,  -6.75,  -5.25,  -3.75,  -2.25,  -1.50  tons. 
Stresses  in  diagonals 

=  +  11.63,  +9.51,  +7.40,  +5.28,  +3.17,  + 1.05  tons. 


50  ROOFS  AND  BRIDGES. 

Prob.  41.    A  through  Warren  truss,  Fig.  26,  has  10  panels, 
each  10  feet  long,  its  web  members  all  forming  equilateral 


triangles  (Art.  14) :  find  the  stresses  in  all  the  web  members 
due  to  a  dead  load  of  400  Ibs.  per  linear  foot  per  truss. 

Ans.  Stresses  in  1-2,  3-4,  5-6,  7-8,  9-10  are  — 10.38, 
-  8.08,  -  5.76,  -  3.46,  -  1.14  tons. 

Stresses  in  1-4,  3-6,  5-8,  7-10,  9-12,  are  + 10.38,  +  8.08, 
+  5.76,  +  3.46,  +  1.14  tons ;  that  is,  the  signs  alternate  in 
the  web  members. 

Art.  19.  Chord  Stresses  due  to  Dead  Loads  — 
Horizontal  Flanges. 

(1)   To  find  the  chord  stresses  by  the  method  of  moments. 

Pass  a  section  cutting  the  chord  member  whose  stress  is 
required,  a  web  member,  and  the  other  chord,  and  take  the 
center  of  moments  at  the  intersection  of  the  web  member 
and  the  other  chord.  Then,  supposing  the  right  part  of  the 
truss  removed,  state  the  equation  of  moments  between  the 
unknown  stress  and  the  exterior  forces  on  the  left  of 
the  section.  For  stresses  in  the  upper  chord  members  the 
centers  of  moments  are  at  the  lower  panel  points;  and  for 
stresses  in  the  lower  chord  members  the  centers  are  at  the 
upper  chord  points. 

Thus,  in  Prob.  41,  each  panel  load  is  2  tons,  and  each 
reaction  is  therefore  9  tons ;  the  depth  of  the  truss  is 
10  sin  60°  =  8.66  feet.  Hence  for  the  lower  chord  stresses, 
we  have : 


BRIDGE  TRUSSES.  51 

9  x  5  -  stress  in  2-4  x  8.66  =  0; 

.-.  Stress  in  2-4  =  +  5.20  tons. 
9x15  —  2x5  —  stress  in  4-6  x  8.66  =  0; 

.-.  Stress  in  4-6  =  +  14.43  tons. 
9  x  25  -  2(15  +  5)  -  stress  in  6-8  x  8.66  =  0; 

.-.  Stress  in  6-8  =  +  21.36  tons. 
9  x  35  -  2  x  45  -  stress  in  8-10  x  8.66  =  0; 

.-.  Stress  in  8-10  =  +  25.98  tons. 
9  x  45  -  2  x  80  -  stress  in  10-12  x  8.66  =  0 ; 

.-.  Stress  in  10-12  =  +  28.28  tons  ; 
and  for  the  upper  chord  stresses,  we  have : 
9  x  10  +  stress  in  1-3  x  8.66  =  0 ; 

.-.  Stress  in  1-3  =  -  10.38  tons. 
9  x  20  -  2  x  10  +  stress  in  3-5  x  8.66  =  0 ; 

.-.  Stress  in  3-5  =  -  18.48  tons. 
9  x  30  -  2(20  +  10)  +  stress  in  5-7  x  8.66  =  0; 

.•.  Stress  in  5—7  =  —  24.24  tons. 
9  x  40  -  2  x  60  +  stress  in  7-9  x  8.66  =  0 ; 

.-.  Stress  in  7-9  =  —  27.70  tons. 
9  x  50  —  2  x  100  +  stress  in  9-9  x  8.66  =  0 ; 

.-.  Stress  in  9-9  =  -  28.86  tons. 

(2)  By  the  method  of  chord  increments. 

Let  it  be  required  to  find  the  stress  in  the  chord  member 
6-8,  Fig.  26.  Denote  the  vertical  shears  in  the  web  mem- 
bers 2-1,  1-4,  4-3,  etc.,  by  vlt  v&  va,  etc.,  and  the  angles 
these  members  make  with  the  vertical  by  01}  02,  03,  etc. 
Now  pass  a  curved  section  cutting  the  chord  member  6-8 
and  all  the  web  members  on  the  left.  Then  from  the  first 


52  ROOFS  AND  BRIDGES. 

condition  of  equilibrium  the  sum  of  the  horizontal  com- 
ponents is  zero.  But  the  horizontal  component  of  the 
stress  in  any  web  member  is  equal  to  the  vertical  shear  in 
that  member  multiplied  by  the  tangent  of  its  angle  with 
the  vertical.  Hence  we  have : 
Stress  in  6-8 

=  Vi  tan  #!  +  v2  tan  62  -f-  v3  tan  03  -f-  vt  tan  B^  +  vs  tan  03. 

Therefore,  to  find  the  stress  in  any  chord  member,  pass  a 
curved  section  cutting  the  member  and  all  the  iveb  members  on 
the  left,  multiply  the  vertical  shear  in  each  web  member  by  the 
tangent  of  its  angle  with  the  vertical,  and  take  the  sum  of  the 
products. 

From  an  inspection  of  Fig.  26  it  is  seen  that  the  stress  in 
any  chord  member,  as  4-6  for  example,  is  greater  than  the 
stress  in  the  immediately  preceding  chord  member,  2-4,  by 
the  sum  of  the  horizontal  components  of  the  stresses  in  the 
web  members,  1-4  and  4-3,  intersecting  at  the  panel  point 
between  these  two  members ;  That  is,  the  increment  of  chord 
stress  at  any  panel  point  is  equal  to  the  sum  of  the  horizontal 
components  of  the  web  stresses  intersecting  at  that  point ;  or, 
equal  to  the  sum  of  the  products  of  the  vertical  shears  of  these 
web  members  by  the  tangents  of  their  respective  angles  with  the 
vertical. 

It  is  well  to  test  the  stress  found  by  this  method  by  the 
method  of  moments,  as  the  results  obtained  by  the  two 
methods  should  agree ;  and  this  affords  a  check  on  the  work. 

The  method  by  chord  increments  applies  only  to  hori- 
zontal chords,  while  the  method  by  moments  applies  to 
trusses  of  any  form,  that  is,  when  the  chords  are  not  hori- 
zontal as  well  as  when  they  are. 

Prob.  42.  Let  it  be  required  to  find  the  chord  stresses  in 
Prob.  41  by  the  method  of  chord  increments. 


BRIDGE   TRUSSES.  53 

Here  each  panel  load  is  2  tons,  and  all  the  web  members 
are  inclined  at  an  angle  of  30° ; 

/.  tan  Oj.  =  tan  02  =  etc.  =  tan  30°  =  .5773. 
Hence  for  the  lower  chord  stresses  we  have : 

Stress  in  2^  =  9  x  .5773  =  +  5.20. 
Stress  in  4-6  =  (9  +  9  +  7)  x  .5773  =  + 14.43. 
Stress  in  6-8  =(9+9+7+7  +  5) x. 5773= +21.36. 
Stress  in  8-10  =(32+5+5+3>x.5773=  +25.98. 
Stress  in  10-12=  (42+3+3  +  1) x. 5773= +28.29; 
and  for  the  upper  chord  stresses  we  have: 

Stress  in  1-3  =  -  (9  +  9)  x  .5773  =  -  10.39. 
Stress  in  3-5  =  -(18+2  x  7)  x  .5773=  -18.47. 
Stress  in  5-7  =  -(32+2  x  5)  x  .5773=  -24.25. 
Stress  in  7-9  =  -(42+2  x  3)  X  .5773  =  -27.71. 
Stress  in  9-9  =  - (48  +2  x  1)  x  .5773  =  - 28.86. 

Prob.  43.  Find  all  the  chord  stresses  in  the  deck  Pratt 
truss  of  Prob.  40. 

Ans.    Stresses  in  lower  chords 

=  +  8.25,  +15.0,  +20.25,  +24.0,  +26.25  tons. 
Stresses  in  upper  chords 

=  _8.25,  -15,  -20.25,  -24,  -26.25,  -27.0  tons. 

Prob.  44.  A  through  Howe  truss,  like  Fig.  25,  has  12 
panels,  each  10  feet  long,  and  10  feet  deep :  find  the  stresses 
in  all  the  members  due  to  a  dead  load  of  400  Ibs.  per  linear 
foot  per  truss. 

Ans.   Lower  chords =11.0,  20.0,  27.0,  32.0,  35.0,  36.0  tons. 
Verticals    =11.0,  9.0,  7.0,  5.0,  3.0,  2.0  tons. 
Diagonals  =15.5,  12.68,  9.86,  7.04,  4.22,  1.4  tons. 
Of  course  the  upper  chord  stresses  can  be  written  directly 
from  those  of  the  lower  chord  by  (1)  of  Art.  5. 


54 


ROOFS  AND  BRIDGES. 


Art.  20.    Position  of  Uniform  Live  Load  Causing 
Maximum  Chord  Stresses. —  Let  Fig.  27  be  a  truss  sup- 


7          9        11         13 


ported  at  the  ends.  Then  to  find  the  stress  in  any  chord 
member,  we  pass  a  section  cutting  that  member,  a  web 
member,  and  the  other  chord,  and  take  the  center  of 
moments  at  the  intersection  of  the  web  member  and  the 
other  chord;  and  since  the  lever  arm  for  the  chord  is  con- 
stant, the  stress  in  any  chord  member  will  be  greatest  when 
the  live  load  is  so  arranged  as  to  give  the  greatest  bending 
moment. 

Now  suppose  we  have  a  uniformly  distributed  moving 
load  coming  on  the  truss  from  the  right,  till  it  produces  the 
left  abutment  reaction  J?x;  then  any  increase  in  the  load  on 
the  right  of  the  section  N  will  affect  the  forces  on  the  left 
only  by  increasing  the  reaction  Rl9  and  consequently  the 
bending  moment.  Hence,  as  R^  increases  with  every  load 
added  to  the  right  of  N,  the  bending  moment  increases,  and 
therefore  the  chord  stress  also  increases.  Also,  suppose  we 
have  a  uniformly  distributed  moving  load  covering  the  truss 
on  the  left  of  the  section  producing  the  right  abutment 
reaction  R2 ;  then  any  increase  in  the  load  on  the  left  of  the 
section  N  will  affect  the  forces  on  the  right  only  by  increas- 
ing the  reaction  Rz,  and  consequently  the  bending  moment. 
Hence  every  load,  whether  on  the  right  or  left  of  the  sec- 
tion, increases  the  bending  moment,  and  therefore  the  chord 
stress. 


BRIDGE   TRUSSES.  55 

Therefore,  for  a  uniform  load,  the  maximum  bending 
moment  at  any  point,  and  consequently  the  maximum  chord 
stress  in  any  member,  occurs  when  the  live  load  covers  the 
whole  length  of  the  truss. 

To  determine  the  chord  stresses  then  due  to  a  uniform 
live  load,  we  have  only  to  suppose  the  live  load  to  cover 
the  whole  truss,  just  as  the  dead  load  does,  and  compute  the 
chord  stresses  in  exactly  the  same  way  as  we  compute  the 
dead  load  stresses. 

Prob.  45.  A  through  Warren  truss,  like  Fig.  26,  has  8 
panels,  each  8  feet  long :  find  the  stresses  in  all  the  chord 
members  due  to  a  live  load  of  1000  Ibs.  per  linear  foot  per 
truss. 

Ans.  Upper  chord  stresses =16. 16,  27.71,  34.64,  36.95  tons. 
Lower  chord  stresses  =  8.08,  21.94,  31.20,  35.80  tons. 

Prob.  46.  A  deck  Pratt  truss  has  11  panels,  each  11  feet 
long,  and  11  feet  deep :  find  all  the  chord  stresses  due  to  a 
live  load  of  800  Ibs.  per  linear  foot  per  truss. 

Ans.   Upper  chord  stresses 

=  22.0,  39.6,  52.8,  61.6,  66.0,  66.0  tons. 

Art.  21.  Maximum  Stresses  in  the  Chords. — Accord- 
ing to  the  principles  of  the  preceding  Article  the  stresses  in 
the  chords  will  be  greatest  when  both  dead  and  live  loads 
cover  the  whole  truss.  We  have  then  only  to  determine 
the  stress  in  each  chord  member  due  to  the  dead  and  live 
loads,  as  in  Arts.  19  and  20,  and  take  their  sum ;  or,  we  may 
add  together  at  first  the  dead  and  live  panel  loads,  and 
determine  the  maximum  stress  in  each  chord  member 
directly.  This  method  is  the  simplest  and  shortest;  but 
it  is  not  the  one  which  in  practice  is  generally  employed. 


56  ROOFS  AND  BRIDGES. 

All  modern  specifications  require  a  separation  of  dead  and  live 
load  stresses.  One  type  of  specification  permits  twice  the  allowable 
stress  per  square  inch  of  metal  for  dead  load  that  it  permits  for  live 
load  ;  thus  10,000  Ibs.  per  square  inch  for  live  load  stresses  and 
20,000  Ibs.  per  square  inch  for  dead  load  stresses.  So  that  if  a  tension 
member  has  a  live  load  stress  of  100,000  Ibs.  and  a  dead  load  stress  of 
60,000  Ibs.,  the  sectional  area  is  determined  as  follows  : 


That  is,  12J  square  inches  are  required  for  the  sectional  area  of  the 
member. 

Another  type  of  specification  requires  the  sectional  area  to  be 
determined  by  a  consideration  of  the  minimum  and  maximum  stresses. 
Thus,  for  tension  members,  the  allowable  stress  is 

10000  (l  +  min.  stress  \lbg  per  gquare  inch 

\        max.  stress  / 
In  the  above  case  the  allowable  stress  would  be 


10000/1  +  -5M™  \  =  13333  ibs.  per  square  inch. 
\          1  50000  y 

Prob.  47.    A  deck  Howe  truss,  for  a  railroad  bridge,  Fig. 
28,  has  12  panels,  each  12  feet  long,  and  .12  feet  deep  ;  the 


8      10     12     lit     12'    10'    8' 
Fig.  38 

dead  load  is  given  by  formula  (4),  Art.  15,  the  live  load  is 
1500  Ibs.  per  linear  foot  per  truss  :  find  the  maximum  chord 
stresses. 

Here  we  have  the  dead  panel  load  per  truss  from  for- 
mula (4) 

=  (6.5x144  +  675)12 

Live  panel  load  per  truss  =  —        —  =9  tons. 
^000 


BRIDGE  TRUSSES.  57 

We  have  then  at  each  upper  apex  a  load  of  5+9=14  tons, 
from  which  the  maximum  chord  stresses  may  be  computed 
directly. 

Ans.   Maximum  stresses  in  lower  chord 

=  77,  140,  189,  224,  245,  252  tons. 

The  upper  chord  stresses  can  be  written  from  the  lower 
by  (1)  of  Art.  5. 

Prob.  48.  A  deck  Howe  truss  has  11  panels,  each  11  feet 
long,  and  11  feet  deep;  the  dead  load  is  400  Ibs.  per  foot 
per  truss,  and  the  live  load  is  1200  Ibs.  per  foot  per  truss : 
find  the  maximum  chord  stresses. 

-4ns.    Lower  chord  stresses  »_ 

=  44.0,  79.2,  105.6,  123.2,  132,  132  tons. 

The  upper  chord  stresses  for  the  Howe  and  Pratt  trusses 
equal  the  lower  chord  stresses  with  contrary  signs,  (1)  of 
Art.  5 ;  also,  all  the  chord  stresses  in  the  Howe  and  Pratt 
are  the  same,  no  matter  on  which  chord  the  loads  may  be 
placed. 

Art.  22.  Position  of  Uniform  Live  Load  Causing 
Maximum  Shears.  — (1)  For  a  uniform  live  load  the 
maximum  positive  shear  at  any  point  N,  Fig.  27,  occurs 
when  every  possible  load  is  added  to  the  right  and  when 
there  is  no  load  on  the  left;  for  the  shear  at  this  point  N  is 
then  equal  to  the  left  reaction  Rlt  and  adding  a  load  to  the 
right  increases  the  left  reaction  Rl  and  therefore  the  posi- 
tive shear,  while  adding  a  load  to  the  left  decreases  the 
positive  shear.  Hence  the  maximum  positive  shear  at  any 
point  occurs  ivhen  the  live  load  extends  from  that  point  to  the 
remote  abutment. 

(2)  Let  a  uniform  live  load  be  placed  on  the  left  of  N, 
producing  the  left  reaction  R^  Now  this  reaction  lil  is 


58  ROOFS  AND  BRIDGES. 

only  a  part  of  the  weight  of  the  live  load,  while  the  shear 
is  this  reaction  minus  the  whole  weight  of  the  load,  and  is 
therefore  negative;  and  every  load  added  to  the  left  of  this 
point  increases  numerically  this  negative  shear.  Hence  the 
maximum  negative  shear  at  any  point  occurs  when  the  live  load 
extends  from  that  point  to  the  nearest  abutment. 

Prob.  49.  Find  the  greatest  positive  and  negative  shears 
for  each  panel  of  Fig.  27  when  the  apex  live  load  is  9  tons. 

By  the  principle  just  proved,  the  greatest  positive  live 
load  shear  in  any  panel  occurs  when  all  joints  on  the  right 
are  loaded,  the  joints  on  the  left  being  unloaded.  Hence, 
for  greatest  positive  shear  in  1-3  the  truss  is  fully  loaded. 
We  have  then 

Positive  shear  in  1-3  =  9  x  2£  =  22.5  tons. 

There  is  no  negative  shear  in  1-3. 

For  greatest  shear  in  3-5  all  joints  except  3  are  loaded. 
Taking  moments  about  right  end,  we  have 

Positive  shear  in       3-5  =  R^  =  9(£  +  £  +  £  +  £ )  =  15  tons. 
Negative  shear  in      3-5  =  Rl  -  9  =  9  x  |  -  9  =  7.5  -  9 

=  —  1.5  tons, 
or,  negative  shear  in  3-5  =  —  9  x  £  =  —  1.5  tons,  as  before. 

Likewise, 

Positive  shear  in       5-7  =  £  (1  +  2  +  3)  =  9  tons. 
Negative  shear  in     5-7  =  —  £  (1  +  2)  =  —  4.5  tons. 
Positive  shear  in       7-9  =  £  (1  +  2)  =  4.5  tons. 
Negative  shear  in      7-9  =  -  £  (1  +  2  +  3)  =  -  9.0  tons. 
Positive  shear  in     9-11  =  £  x  1  =  1.5  tons. 
Negative  shear  in    9-11  =  —  $(1  +  2  +  3  +  4)=  — 16  tons. 
Negative  shear  in  11-13  =  -  £  (1  +  2  +  3  +  4  +  5) 

=  -  22.5  tons. 


BRIDGE   TRUSSES. 


59 


There  is  no  positive  shear  in  11-13. 

The  greatest  positive  shears  are  equal  and  symmetrical  to 
the  greatest  negative  shears.  Thus,  the  positive  shear  in 
1-3  is  equal  numerically  to  the  negative  shear  in  11-13; 
the  positive  shear  in  3-5  is  equal  numerically  to  the  nega- 
tive shear  in  9-11 ;  and  so  on. 

Prob.  50.  Find  the  greatest  positive  and  negative  shears 
for  each  panel  of  Fig.  25  when  the  apex  live  load  is  6  tons. 

Ana.  Positive  shears  =  21.0,  15.75,  11.25,  7.5,  4.5,  2.25, 
0.75,  0.0  tons. 

Negative  shears  =  0.0,  0.75,  2.25,  4.5,  7.5,  11.25,  15.75, 
21.0  tons. 

Prob.  51.  Find  the  maximum  positive  and  negative  shears 
for  each  panel  of  Fig.  28  when  the  apex  dead  and  live  loads 
are  5  and  9  tons,  respectively,  as  in  Prob.  47. 

Ans. 


MEMBERS. 

2-4 

4-6 

6-8 

8-10 

10-12 

12-14 

+27.50 

+22.50 

+17.50 

+12.50 

+  750 

+  2.50 

Live  load  positive  shear     .    . 
Live  load  negative  shear    .    . 

+49.50 
0.00 

+41.25 
-  0.75 

+33.75 
-  2.25 

+27.00 
-  4.50 

+21.00 
-  7.60 

+15.75 
-11.25 

+77.00 

+63.75 

+51.25 

+39.50 

+28.50 

+18.25 

Minimum  shear    

+27.50 

+21.75 

+15.25 

+  8.00 

0.00 

-  8.75 

The  maximum  shear  is  always  positive  in  the  left  half  of 
the  truss,  but  the  minimum  may  be  either  positive  or  nega- 
tive. In  this  problem  we  see  that  a  negative  shear  can 
occur  in  panel  12-14,  and  that  there  is  no  shear  in  panel 
10-12,  while  the  shears  in  all  the  other  panels  are  positive. 
Members  to  the  left  of  10,  and  of  course  in  the  four  right 
hand  panels  also,  are  therefore  subjected  to  but  one  kind  of 
stress,  while  in  the  four  middle  panels  they  may  be  sub- 


60  ROOFS  AND  BRIDGES. 

jected  to  either  kind  of  stress,  and  hence  must  be  counter- 
braced.  Though  there  is  no  negative  shear  in  panel  10-12, 
it  would  be  counterbraced  for  safety. 

Art.  23.  The  Warren  Truss.  —  The  chord  stresses  in 
the  Warren  Truss  may  be  computed  by  the  method  of  chord 
increments,  (2)  of  Art.  19,  and  the  web  stresses  by  the 
method  of  Arts.  18  and  22.  The  dead  and  live  load  stresses 
may  be  found  separately  and  their  sum  taken  for  the  maxi- 
mum and  minimum  stresses;  or  the  maximum  and  mini- 
mum stresses  of  any  member  may  be  determined  directly, 
from  a  single  equation,  by  placing  the  dead  and  live  loads 
in  proper  position.  In  the  following  solution  of  the  deck 
Warren  truss  the  dead  and  live  load  stresses  are  found 
separately. 

Prob.  52.  A  deck  Warren  truss,  as  a  deck  railroad  bridge, 
Fig.  29,  has  10  panels,  each  12  feet  long,  its  web  members 


Fig.  3D 

all  forming  equilateral  triangles  (Art.  14)  ;  the  dead  load  is 
given  by  formula  (2),  Art.  15,  the  live  load  is  1500  Ibs.  per 
foot  per  truss  :  find  the  maximum  and  minimum  stresses  in 
all  the  members. 

The  dead  panel  load  per  truss  from  formula  (2) 

=  (7x120  +  600)12  m  8640  lbs<  =  4.32  tons 
=  say,  4  tons,  for  convenience  in  computation. 


Live  panel  load  per  truss  =  =  9  tons. 


BEIDGE  TRUSSES.  61 

Since  the  loads  are  distributed  uniformly  along  the  joints 
of  the  upper  chord,  3  and  3'  will  receive  three  fourths  of  a 
panel  load  each.  The  dead  and  live  load  stresses  will  be 
found  separately,  tan  6  =  .577 ;  sec  0  =  1.1547. 


BEAD  LOAD  STRESSES. 

Left  reaction  =  4.75  x  4  =  19  tons.  This  is  also  the  shear 
in  panel  1-3.  The  dead  load  shear  in  each  of  the  other 
panels  is  found  by  subtracting  the  loads  on  the  left  of  the 
panel  from  the  abutment  reaction.  We  have  then  the  fol- 
lowing chord  stresses  by  chord  increments : 


UPPER  CHORD  STRESSES. 

Stress  in      3-5  =  (19  +  16)  x  .577  =  -  20.20  tons. 

Stress  in      5-7  =  (19  +  2  x  16  +  12)  x  .577  =  -  36.36  tons. 
Stress  in      7-9  =  (19  +  2  x  16  +  2  x  12  +  8)  x  .577 

=  -  47.90  tons. 

Stress  in    9-11  =  (75  +  2  x  8  +  4)  x  .577     =  -  54.82  tons. 
Stress  in  11-11'=  (91  +  2  x  4)  x  .577  =  -  57.12  tons. 

Stress  in      1-3  =     00.00  tons. 

LOWER  CHORD  STRESSES. 

Stress  in      2-4  =  19  x  .577  =  10.96  tons. 

Stress  in      4-6  =  (19  +  2  x  16)  x  .577  =  29.42  tons. 
Stress  in      6-8  =  (51  +  2  x  12)  x  .577  =  43.28  tons. 
Stress  in    8-10  =  (75  +  2  x  8)  x  .577    =  52.51  tons. 
'  Stress  in  10-12  =  (91  +  2  x  4)  x  .577    =  57.12  tons. 


62  ROOFS  AND  BRIDGES. 

WEB  STRESSES  (BY  ART.  18). 

Stress  in      2-3=19  x  1.154=  -21.92  tons. 
Stress  in      3-4 =16x1.154  =+18.46  tons  =- stress  in  4-5. 
Stress  in      5-6  =  12  x  1.154=  +13.84  tons  =  -stress  in  6-7. 
Stress  in      7-8=  8x1.154=+  9.23  tons  =— stress  in  8-9. 
Stress  in    9-10=  4x1.154=+  4.61  tons 

=  —  stress  in  10-11. 

Stress  in  11-12=0.00          =  -stress  in  12-11' 
Stress  in      1-2 =£  of  a  panel  load=l  ton. 

LIVE  LOAD  STRESSES. 

The  maximum  live  load  chord  stresses  occur  when  the 
live  load  covers  the  whole  length  of  the  truss  (Art.  20); 
hence  they  are  found  in  precisely  the  same  way  as  the  dead 
load  chord  stresses  above ;  or  we  may  find  them  by  multi- 
plying the  above  dead  load  chord  stresses  by  the  ratio  of 
live  to  dead  panel  load,  which  =  2\  here,  giving  us  the 
following  chord  stresses : 

UPPER  CHORD  STRESSES.  LOWER  CHORD  STRESSES. 

1-3  =       00.00  tons.  2-4  =    24.66  tons. 

3-5  =  -   45.45  tons.  4-6  =    66.19  tons. 

5-7  =  -    81.81  tons.  6-8  =    97.38  tons. 

7_9  =  _  107.77  tons.  8-10  =  118.15  tons. 

9-11  =  -  123.34  tons.  10-12  =  128.52  tons, 

11-11 '=-128.52  tons. 


BRIDGE  TRUSSES.  63 

WEB  STRESSES. 

The  stress  in  any  web  member  is  equal  to  the  shear  in 
the  section  which  cuts  that  member  and  two  horizontal  chord 
members,  multiplied  by  the  secant  of  the  angle  which  the 
web  member  makes  with  the  vertical  (Art.  18)  ;  the  stress 
is  therefore  a  maximum  when  the  shear  is  a  maximum. 
The  maximum  positive  live  load  shear  in  any  panel  occurs 
when  all  joints  on  the  right  are  loaded  and  the  joints  on 
the  left  are  unloaded  (Art.  22). 

The  maximum  positive  live  load  shear  for  2-3  will  occur 
when  the  truss  is  fully  loaded.  We  have  then 

Stress  in  2-3  =  9  x  4.75  x  1.154  =  -  49.30  tons. 

The  maximum  positive  shear  for  3-4  and  4-5  will  occur 
when  all  joints  except  3  are  loaded  (Art.  22).  Taking 
moments  about  the  right  end,  and  multiplying  by  sec  0,  we 
have 

Stress  in  3-4 

=  ^  (f  +  3  +  5  +  7  +  9  +  11  +  13  +  15  -(-  17)  x  1.1547 
=  ^  x  323  x  Iil547  =  41>96  tons  =  _  stress  in  4_5i 

Stress  in  5-6 


=  33.13  tons  =  -  stress  in  6-7. 
Stress  in      7-8  =  &  (f  +  3  +  5+7+94-11+13)  x  1.1547 

=  25.33  tons  =  —  stress  in  8-9. 
Stress  in    9-10  =  -&  (f  +3  +  5  +  7  +  9  +  11)  x  1.1547 

=  18.58  tons  =  -  stress  in  10-11. 
Stress  in  11-12  =  ^  (£  +  3  +  5  +  7  +  9)  x  1.1547 

=  12.87  tons  =  -  stress  in  12-11'. 


64 


ROOFS  AND  BRIDGES. 


Stress  in  11 '-10'  =  ^(f  +  3  +  5  +  7)  x  1.1547 

=  8.19  tons  =  —  stress  in  10 '-9'. 
Stress  in      9'-8'  =  ^  (f  +  3  +  5)  x  1.1547 

=  4.55  tons  =  —  stress  in  8 '-7'. 
Stress  in      7'-6'  =  ^ (|  +  3)  x  1.1547 

=  1.96  tons  =  —  stress  in  6 '-5'. 
Stress  in      5'-4'  =  -&  (f)  X  1.1547 

=  0.40  tons  =  -  stress  in  4'-3'. 
Stress  in      3'-2'  =  0.00  tons. 

Collecting  the  above  results  we  may  enter  them  in  a  table 
as  follows : 

TABLE  OF  STKESSES  IN  ONE  TRUSS. 
UPPER  CHORD  STRESSES. 


MEMBERS. 

1-3 

3-5 

5-7 

T-9 

9-11 

11-11' 

Dead  load 
Live  load 

00.00 
00.00 

-20.20 
-46.45 

-  36.36 
-  81.81 

-  47.90 
-107.77 

-  54.82 
-123.34 

-  57.12 
-128.52 

Max.  stress 
Min.  stress 

00.00 
00.00 

-65.65 
-20.20 

-118.17 
-  36.36 

-155.67 
-  47.90 

-178.16 
-  54.82 

-185.64 
-  57.12 

LOWER  CHORD  STRESSES. 


MEMBEBS. 

2-1 

4-6 

6-S 

8-10 

10-12 

Dead  load   .  . 
Live  load  .  .  . 

+  10.96 
+24.66 

+29.42 
+66.19 

+  43.28 
+  97.38 

+  52.51 
+  118.15 

+  57.12 
+  128.52 

Max.  stress  .  . 
Min.  stress  .  . 

+35.62 
+  10.96 

+  95.61 
+29.42 

+  140.66 
+  43.28 

+  170.66 
+  52.61 

+  185.64 
+  57.12 

BRIDGE  TRUSSES. 


65 


WEB  STRESSES. 


MEMBERS. 

2-3 

W 

4-5 

5-6 

6-7 

^  Dead  load    .... 

-21.85 

+  18.46 

-18.46 

+  13.84 

-13.84 

J  \  From  right  .  .  . 

-49.30 

+41.96 

-41.96 

+  33.13 

-33.13 

.>  \  From  left  .... 

00.00 

-  0.40 

+  0.40 

-   1.96 

+   1  96 

Max  stress  

-71.15 

+  60.42 

-60.42 

+4697 

-4697 

Min.  stress  

-21.85 

+  18.06 

-18.06 

+  11.88 

-11.88 

MEMBERS. 

7-8 

8-9 

9-10 

10-11 

11-12 

•  Dead  load    .... 

+  9.23 

-  9.23 

+  4.61 

-  4.61 

+  0.00 

J  j  From  right  .  .  . 

+  25.33 

-25.33 

+  18.58 

-18.58 

+  12.87 

|  (  From  left  .... 

-  4.55 

+  4.65 

-  8.19 

+  8.19 

-12.87 

Max.  stress  

+  34.56 

-34.56 

+23.19 

-23.19 

+  12.87 

Min.  stress  

+  4.68 

-  4.68 

-  3.58 

+  3.58 

-  12.87 

"We  see  from  this  table  of  web  stresses  that,  (1)  when  the 
live  load  conies  on  from  the  right,  all  the  web  members  in 
the  left  half  of  the  truss  are  subjected  to  but  one  kind  of 
stress,  that  is,  the  members  2-3,  4-5,  6-7,  8-9,  10-11  are 
subjected  to  compressive  stresses,  and  the  members  3-4,  5-6, 
7-8,  9-10,  11-12  are  subjected  to  tensile  stresses ;  (2)  when 
the  live  load  comes  on  from  the  left,  all  the  web  members 
to  the  left  of  9-10  are  subjected  to  but  one  kind  of  stress, 
but  the  members  9-10,  10-11,  11-12  have  their  stresses 
changed,  that  is,  the  stresses  in  9-10  and  11-12  are  changed 
from  tensile  to  compressive,  while  that  in  10-11  is  changed 
from  compressive  to  tensile. 

^     Hence,  the  members  2-3,  4-5,  6-7,  8-9,  should  be  struts 
to  carry  compression  only,  and  the  members  3—4,  5-6,  7-8, 


66  ROOFS  AND  BRIDGES. 

should  be  ties  to  carry  tension  only,  while  9-10,  10-11, 
11-12  should  be  members  capable  of  resisting  both  com- 
pression and  tension,  and  should  therefore  be  counter  braces 
(Art.  1) :  and  the  same  is  true  for  the  right  half  of  the  truss. 

In  this  solution  for  web  stresses  the  live  load  was  brought 
on  from  the  right,  and  the  maximum  positive  shear  was 
found  in  each  panel  of  the  right  half  of  the  truss,  and  then 
the  resulting  stress;  if  preferred,  the  live  load  may  be 
brought  on  from  the  left,  and  the  maximum  negative  shear 
be  found  in  each  panel  of  the  left  half  of  the  truss  (Art.  22), 
and  then  the  resulting  stress.  Thus,  the  maximum  positive 
shear  in  any  panel  in  the  right  half  of  the  truss  as  6 '-8'  has 
the  same  numerical  value  as  the  maximum  negative  shear 
in  the  corresponding  member  6-8  in  the  left  half ;  and  the 
resulting  stress  in  any  member  7'-8'  has  the  same  value  as 
that  in  the  corresponding  member  7-8. 

The  maximum  and  minimum  stresses  may  be  determined 
directly  from  a  single  equation  by  placing  the  dead  and 
live  loads  in  proper  position. 

Thus,  for  the  maximum  stress  in  8-9,  we  pass  a  section 
cutting  it,  place  the  live  load  on  the  right,  and  have 

Max.  stress  in  8-9= -[2x4+^(f +3+5+7  +  9+11+13)] 
x  1.1547=— 34.57  tons,  as  before. 

For  the  minimum  stress  the  live  load  is  reversed  and 
covers  the  truss  on  the  left  of  the  section.  Thus 

Min.  stress  in  8-9= -[2x4-^(1 +3+5)]  x  1.1547 
=  —4.69  tons,  as  before. 

Prob.  53.  A  deck  Warren  truss,  like  Fig.  29,  has  10 
panels,  each  10  feet  long,  its  web  members  all  forming  equi- 
lateral triangles ;  the  dead  load  is  800  Ibs.  per  foot  per 
truss,  and  the  live  load  is  1600  Ibs.  per  foot  per  truss ;  the 


BRIDGE  TRUSSES. 


67 


joints  3  and  3'  receive  three  fourths  of  a  panel  load  each : 
find  the  maximum  and  minimum  stresses  in  all  the  members. 


Ans. 


UPPER  CHORD  STRESSES. 


MEMBERS. 

1-3 

3-5 

5-7 

7-9 

9-H 

ii-ii' 

Max.  stresses 

00.00 

-60.60 

-109.08 

-143.76 

-164.52 

-171.48 

Min.  stresses 

00.00 

-20.20 

-  86.36 

-  47.92 

-  54.84 

-  57.16 

LOWER  CHORD  STRESSES. 


MEMBERS. 

2-4 

4-6 

6-8 

8-10 

10-12 

Max.  stresses  .  .  . 
Min.  stresses    .  .  . 

+32.88 
+  10.96 

+88.20 
+  29.40 

+  129.84 
+  43.28 

+  157.56 
+  52.52 

+  171.48 
+  57.16 

WEB  STRESSES. 


MEMBERS. 

2-3 

3-4 

4-5 

5-6 

6-7 

Max.  stresses  .  .  . 
Min.  stresses   .  .  . 

-65.82 
-21.84 

+  55.80 
+  18.12 

-55.80 
-18.12 

+43.29 
+  12.09 

-43.29 
-12.09 

MEMBERS. 

7-8 

8-9 

9-10 

10-11 

11-12 

Max.  stresses  .  .  . 
Min.  stresses    .  .  . 

+31.78 
+  5.18 

-31.78 

-  5.18 

+21.13 
-  2.70 

-21.13 
+  2.70 

+  11.46 
-11.46 

Prob.  54.  A  deck  Warren  truss,  like  Fig.  29,  has  8  panels, 
each  15  feet  long,  its  web  members  all  forming  equilateral 
triangles;  the  dead  load  is  given  by  formula  (2)  of  Art.  15, 
the  live  load  is  1600  Ibs.  per  foot  per  truss,  the  joints  3  and 
3'  receive  three  fourths  of  a  panel  load  each :  find  the  maxi- 
mum and  minimum  stresses  in  all  the  members. 


68  ROOFS  AND  BRIDGES. 

Aiis.  Max.  stresses  in  upper  chord 

=  0.00,  -67.77,  -117.97,  -148.09,  -15^.17  tons. 
Min.  stresses  in  upper  chord 

=0.00,  -21.00,  -36.61,  -46.01,  -49.09  tons. 
Max.  stresses  in  lower  chord 

=  +37.65,  +97.89,  +138.05,  +158.13  tons. 
Min.  stresses  in  lower  chord 

=  +11.66,  +30.40,  +42.87,  +49.09  tons. 
Max.  web  stresses 

=  -75.35,  +60.92,  -60.92,  +43.43,  -43.43,  +27.67, 

-27.67,  +13.64  tons. 
Min.  web  stresses 

=  -23.38,  +18.06,    -18.06,    +9.23,  —9.23,   -  !.:*>. 
+  1.35,  -13.64  tons. 

Prob.  55.  A  through  Warren  truss,  Fig.  26,  has  10  panels, 
each  12  feet  long,  its  web  members  all  forming  equilateral 
triangles;  the  dead  load  is  500  Ibs.  per  foot  per  truss,  and 
the  live  load  is  834  Ibs.  per  foot  per  truss :  find  the  maxi- 
mum and  minimum  stresses  in  all  the  members. 

Ans.   Upper  chord  max. 

=  -41.52,  -73.92,  -96.96,  -110.80,  -115.44  tons. 

Upper  chord  min. 

=  -15.57,  -27.72,  -36.36,  -41.55,  -43.29  tons. 

Lower  chord  max. 

=  +  20.80,  +57.76,  +85.44,  +103.92,  +113.12  tons. 

Lower  chord  min. 

=  +  7.80,  +  21.66,  +  32.04,  +  38.97,  +  42.42  tons. 
Web  stresses  max. 

=  -41.55,  +41.55,  -32.91,  +32.91,  -24.81,  +24.81, 
-  17.31;  +17.31,  -  10.37,  +  10.37  tons. 


BRIDGE  TRUSSES.  t)9 

Web  stresses  min. 

=  -  15.57,  + 15.57,  -  11.55,  +  11.55,  -  6.91,  +  6.91, 

-  1.73,  +  1.73,  +  4.06,  -  4.06  tons. 
Prob.  56.   A  through  Warren  truss,  like  Fig.  26,  has  8 
panels,  each  10  feet  long,  its  web  members  all  forming  equi- 
lateral triangles ;  the  dead  load  is  800  Ibs.  per  foot  per  truss, 
and  the  live  load  1600  Ibs.  per  foot  per  truss :  find  the  maxi- 
mum and  minimum  stresses  in  all  the  members. 
Ans.  Upper  chord  max. 

=  -48.48,  -83.10,  -103.86,  -110.79  tons. 
Upper  chord  min. 

=  -16.16,  -27.70,  -34.62,  -36.93  tons 
Lower  chord  max. 

=  +24.24,  +65.76,  +93.48,  +107.31  tons. 
Lower  chord  min. 

=  +8.08,  +21.92,  +31.16,  +35.77  tons. 
"Web  sfresses  max. 

=  -48.48,  +48.48,  -35.77,  +35.77,  -24.23,  +24.23, 

—13.85,  + 13.85  tons. 
Web  stresses  min. 

=  -16.16,  +16.16,  -10.40,  +10.40,  -3.46,  +3.46, 
+  4.61,  —4.61  tons. 

Prob.  57.  A  deck  Warren  truss,  like  Fig.  29,  has  7  panels, 
each  15  feet  long  and  15  feet  deep,  its  web  members  all 
forming  isosceles  triangles ;  the  dead  load  is  4  tons,  and  the 
live  load  is  9  tons,  per  panel  per  truss,  the  first  and  last 
joints  3  and  3'  receiving  three  quarters  of  a  panel  load  each, 
as  in  Probs.  52,  53,  and  54 :  find  the  maximum  and  mini- 
mum stresses  in  all  the  members. 

Ans.  Upper  chord  max. 

=  _  o.OO,  -  37.38,  -  63.38,  -  76.38  tons. 


70  ROOFS  AND  BRIDGES. 

Upper  chord  min. 

=  _  0.00,  -  11.50,  -  19.50,  -  23.50  tons. 
Lower  chord  max. 

=  +  21.13,  +  53.65,  +  73.13,  +  79.63  tons. 
Lower  chord  min. 

=  +  6.50,  +  16.50,  +  22.50,  +  24.50  tons. 
Web  stresses  max. 

=  -47.23,  +36.87,  -36.87,  +24.49,  -24.49,  +13.56, 

- 13.56  tons. 
Web  stresses  min. 

=  -  14.53,   +  10.64,  -  10.64,   +  4.01,  -  4.01,  -  4.05, 
+  4.05  tons. 

Art.  24.  Mains  and  Counters.  —  In  the  Pratt  truss, 
Fig.  30,  all  the  verticals,  except  1-4  and  l'-4',  are  to  take 
compression  only,  and  all  the  inclined  members, -except  1-2, 


1        3        5       5'       3'      1' 


1-2'  are  to  take  tension  only.  The  two  inclined  members, 
1-2  and  l'-2',  are  usually  called  the  "inclined  end  posts." 
The  verticals  1-4  and  l'-4'  do  not  form  any  part  of  the 
truss  proper,  since  they  serve  only  to  carry  the  loads  at  4 
and  4'  up  to  the  hip  joints  1  and  1';  they  are  called  hip 
verticals. 

The  members  1-6,  3-8,  3'-8f,  and  l'-6'  are  mains,  or  main 
ties.  They  are  the  only  inclined  ties  that  are  put  under 
stress  by  the  action  of  the  dead  load,  or  by  a  uniform  dead 
and  live  load  extending  over  the  whole  truss.  The  mem- 
bers 5-6,  5-8',  5'-8,  and  5 '-6',  in  dotted  lines,  are  counters, 


BRIDGE  TRUSSES.  71 

or  counter  ties.  In  the  Pratt  truss  there  are  no  counter 
braces  (Art.  1).  These  four  counters  are  called  into  play 
only  when  the  live  load  covers  a  part  of  the  truss. 

Thus,  if  a  load  be  placed  at  the  joint  6',  it  is  carried  by 
the  truss  to  the  abutments  at  2  and  2'.  The  part  of  this 
load  which  goes  to  2  may  be  conceived  as  being  carried  up 
to  5',  down  to  8',  up  to  5,  down  to  8,  up  to  3,  down  to  6,  up  to 

1,  down  to  the  abutment  at  2.     The  other  part  of  this  load, 
which  goes  to  the  right,  passes  up  to  1',  then  down  to  2'. 
For  this  loading,  the  counters  5'-6',  5-8',  and  the  mains  3-8, 
1-6,  and  1-6'  are  put  under  stress,  while  the  counters  5'-8, 
5-6,  and  the  main  tie  3'-8'  are  idle,  and  might  be  removed 
Avithout  endangering  the  truss.     If  the  two  middle  joints, 
8  and  8',  are  loaded  equally,  the  part  of  the  load  at  8'  going 
to  2'  is  just  balanced  by  the  part  of  the  load  at  8  going  to 

2,  and  hence  there  is  no  stress  in  the  intermediate  web 
members  5-8',  5'-8,  5-8,  and  5'-8',  nor  in  the  counters  5-6 
and  5'-6'.     If  the  live  load  going  from  6'  to  the  left  abut- 
ment is  greater  than  the  dead  load  going  from  8'  to  the 
right  abutment,  the  counter  5'-6'  must  be  inserted.     If  this 
counter  5'-6'  were  not  inserted,  the  panel  6'-8'  would  be 
distorted.     The  load  at  6'  would  bring  the  opposite  corners 
3'  and  8'  nearer  together,  and  the  opposite  corners  5'  and 
6'  farther  apart,  because  the  main  tie  3'-8'  cannot  take  com- 
pression.    Whenever  the  live  load  would  tend  to  cause  com- 
pression in  any  main  tie,  a  counter  must  be  inserted  uniting 
the  other  corners  of  the  panel.     Both  diagonals  in  any  panel 
cannot  have  compression  or  tension  at  the  same  time. 

If  the  action  of  the  dead  and  live  loads  tend  to  subject  a 
member  to  stresses  of  opposite  kinds,  the  resultant  stress 
in  the  member  will  be  equal  to  the  numerical  difference  of 
the  opposite  stresses,  and  will  be  of  the  same  kind  as  the 
greater.  Thus,  if  the  dead  load,  going  from  8  to  the  left 


72  ROOFS  AND  BRIDGES. 

abutment,  should  subject  the  main  tie  3-8  to  a  tension  of  8 
tons,  and  if  the  live  load,  going  from  6  to  the  right  abut- 
ment, should  tend  to  subject  the  same  tie  to  a  compression 
of  5  tons,  there  would  be  a  resulting  tension  of  3  tons  in 
the  member  3-8 ;  and  the  counter  5-6  would  not  be  needed 
for  this  loading.  But,  if  the  dead  load  should  subject  the 
tie  3-8  to  a  tension  of  only  2  tons,  while  the  the  live  load 
should  tend  to  subject  the  same  piece  to  a  compression  of 
5  tons,  there  would  be  a  resulting  force  of  3  tons,  tending 
to  bring  the  joints  3  and  8  nearer  together,  or  to  separate 
the  joints  5  and  6  from  each  other.  In  order  to  provide  for 
this  compression  or  tension,  either  a  brace  would  be  needed 
alongside  the  main  tie  3-8,  to  take  the  compression  of  3  tons, 
or  the  counter  tie  5-6  would  be  needed  to  take  the  tension 
of  3  tons. 

In  the  Howe  truss,  Fig.  31,  the  verticals  are  to  take  ten- 
sion only,  and  the  inclined  members  are  to  take  compression 


2         It        6        8        8'      6' 
Fig.  31 

only.  The  mains,  or  main  braces,  are  represented  by  the 
full  lines ;  the  dotted  lines  denote  counters,  or  counter 
braces.  Under  the  action  of  the  dead  load,  or  of  a  uniform 
dead  and  live  load  covering  the  whole  truss,  the  main 
braces  are  the  only  diagonals  that  are  put  under  stress. 
The  counter  braces  are  called  into  play  only  when  the  live 
load  covers  a  part  of  the  truss,  as  in  the  case  of  the  Pratt 
truss. 

Thus,  if  a  load  be  placed  at  the  joint  6,  the  part  of  it 
which  goes  to  the  right  abutment  2'  may  be  conceived  as 


BRIDGE   TRUSSES.  73 

being  carried  up  to  3,  down  to  8,  up  to  5,  down  to  8',  up  to 
5',  down  to  6',  up  to  3',  down  to  4',  up  to  1',  down  to  2'. 
If  the  live  load  going  from  6  to  the  right  abutment  is 
greater  than  the  dead  load  going  from  8  to  the  left  abut- 
ment, the  counter  brace  3-8  must  be  inserted.  If  this 
counter  3-8  were  not  inserted,  the  panel  6-8  would  be  dis- 
torted. The  load  at  6  would  bring  the  opposite  corners  3 
and  8  nearer  together,  and  the  opposite  corners  5  and  6 
farther  apart,  because  the  main  brace  5-6  cannot  take 
tension.  But  if  the  live  load  going  from  6  to  the  right 
abutment  is  less  than  the  dead  load  going  from  8  to  the  left 
abutment,  the  counter  brace  3-8  will  not  be  needed  for  this 
loading.  The  main  and  counter  in  any  panel  cannot  both 
take  stress  at  the  same  time  by  any  system  of  loading. 

We  may  determine  where  the  counters  are  to  begin,  as 
follows:  Consider  the  left  half  of  the  truss,  and  let  the 
live  load  come  on  from  the  left ;  then  the  dead  and  live  load 
shears,  or  stresses,  are  of  opposite  kinds.  It  results  from 
this  at  once  that  counters  —  counter  ties  in  the  Pratt,  and 
counter  braces  in  the  Howe  —  must  begin  in  that  panel  in 
which  the  stress  or  shear  caused  by  the  live  load  is  greater 
than  that  of  the  opposite  kind  caused  by  the  dead  load. 

Thus,  in  Fig.  30,  let  the  dead  panel  load  be  3  tons  and 
the  live  panel  load  14  tons.  Then  we  have  the  following 
maximum  negative  shears : 

Shear  in  2^t  =  +  9  -  0  =  +  9. 

Shear  in  4-6  =  +  6  -  -f  x  14        =  +  4. 
Shear  in  6-8  =  +  3  -  14  (|  +  |)  =  -  3. 

Hence  the  counters  must  begin  in  the  third  panel,  and 
therefore  the  third,  fourth,  and  fifth  panels  must  have 
counter  ties  for  this  loading. 


74 


ROOFS  AND  BRIDGES. 


Prob.  58.  In  the  Howe  truss  of  Fig.  28,  with  12  panels, 
the  dead  panel  load  being  4  tons  and  the  live  9  tons,  find 
the  number  of  panels  to  be  counterbraced. 

Ans.  The  5th,  6th,  7th,  and  8th  panels  must  have  counter 
braces. 

Art.  25.  The  Howe  Truss.  —  The  maximum  and  mini- 
mum stresses,  both  in  the  chords  and  the  web  members  of 
the  Howe  truss,  may  now  be  computed  by  the  principles  of 
Arts.  18,  19,  22,  and  24.  The  dead  and  live  load  stresses 
may  be  found  separately,  and  their  sum  taken  for  the  maxi- 
mum and  minimum  stresses ;  or,  the  maximum  and  mini- 
mum stresses  of  any  member  may  be  determined  directly, 
from  a  single  equation,  by  placing  the  dead  and  live  loads 
in  proper  position.  In  the  following  solution  of  the  through 
Howe  truss,  this  method  is  the  one  employed : 

Prob.  59.  A  through  Howe  truss,  as  a  through  railroad 
bridge,  Fig.  32,  has  10  panels,  each  12  feet  long  and  12  feet 


deep ;  the  dead  load  is  given  by  formula  (4),  Art.  15,  the 
live  load  is  1500  Ibs.  per  foot  per  truss :  find  the  maximum 
and  minimum  stresses  in  all  the  members. 

From  formula  (4),  Art.  15,  we  find  the  dead  panel  load 
per  truss 

=  (6.5x120+675)12  =  873Q  ^ 


=  say,  4  tons,  for  convenience  of  computation. 

Live  panel  load  per  truss  =  15(ff*  12  =  9  tons. 
2000 

tan  0  =  1,  sec  d  =  1.41. 


BRIDGE  TRUSSES.  75 

The  maximum  and  minimum  chord  stresses  may  be  written 
out  at  once  (Arts.  19  and  21).  They  are  as  follows : 

Max.  stresses  in  lower  chord  =  58.5,  104.0,  136.5,  156.0, 
162.5  tons. 

Min.  stresses  in  lower  chord= 18.0, 32.0, 42.0, 48.0,50.0  tons. 

The  upper  chord  stresses  may  be  written  from  the  lower 
at  once.  Thus,  stress  in  3-5  =  —  stress  in  1-A ;  and 
so  on. 

WEB  STRESSES. 

The  maximum  stress  in  any  web  member  is  equal  to  the 
maximum  shear  in  the  section  which  cuts  that  member  and 
two  horizontal  chord  members,  multiplied  by  the  secant  of 
the  angle  which  the  web  member  makes  with  the  vertical 
(Art.  18).  The  maximum  positive  live  load  shear  in  any 
panel  in  the  left  half  of  the  truss  occurs  when  all  joints  on 
the  right  are  loaded  and  the  joints  on  the  left  are  unloaded 
(Art.  22).  For  the  maximum  stress  in  2-3,  we  pass  a  sec- 
tion cutting  it,  place  the  live  load  on  the  right,  in  this  case 
covering  the  whole  truss,  and  have 

Max.  stress  in  2-3 

=  -  [4  x  4.5  +  -^(1+2  +  3  +  4  +  5  +  6  +  7+8  +  9)]  1.41 
=  -  82.48  tons. 

For  the  max.  stress  in  4-5  all  the  joints  on  the  right  of 
4  are  loaded.  Hence, 

Max.  stress  in  4-5 
=  _[4x3.5+T<V(l  +  2+3+4  +  •••+  8)]  1.41  =-65.42  tons. 

Max.  stress  in  6-7 
=  -  [4  x  2.5  +  -& (1  +  2  +  3  +  •••  +  7)]  1.41  =  -  49.63 tons. 

Max.  stress  in  8-9 
==-[4  x  1.5  +  ^(1+2  +  3+  .-+6)]  1.41  =-35.10  tons. 

Max.  stress  in  10-11 
=  -  [4  x  0.5  +  -ft(l  +  2  +  -  +  5)]  1.41  =  -  21.85  tons. 


76  ROOFS  AND  BRIDGES. 

For  the  minimum  stress  in  any  member  in  the  left  half 
of  the  truss,  the  live  load  is  reversed  and  covers  the  truss 
on  the  left  of  the  section  (Art.  22).  Hence, 

Min.  stress  in  2-3=  -[4  x  4.5  -  0]  1.41  =  -  25.38  tons. 
Min.  stress  in  4-5  =  -  [4  x  3.5  -  ^  x  1]  1.41  =  - 18.47  tons. 
Min.  stress  in  6-7 

=  -  [4  x  2.5  -  ^  (1  +  2)]  1.41  =  -  10.29  tons. 
Min.  stress  in  8-9 

=  -  [4  x  1-5  -  T\  (1  +  2  +  3)]  1.41  =  -  0.85  tons. 
Min.  stress  in  10-11 

=  -  [4  x  0.5  -  ^  (1  +  2  +  3  +  4)]  1.41  =  +  9.87  tons. 

That  is,  the  minimum  stress  in  10-11  is  a  tension  of  9.87 
tons.  But  as  the  diagonals  in  the  Howe  truss  are  to  be  sub- 
jected only  to  compression,  the  brace  10-11  cannot  take  ten- 
sion, the  counter  brace  9-12  must  therefore  be  inserted  to 
take  this  9.87  tons,  which  is  the  excess  of  that  part  of  the 
live  load  going  from  10  to  the  right  abutment  over  that  part 
of  the  dead  load  going  from  12  to  the  left  abutment.  If  a 
tie  were  placed  alongside  the  main  brace  10-11,  it  would 
take  this  tension  of  9.87  tons,  and  the  counter  brace  9-12 
would  not  be  needed ;  or,  if  the  diagonal  10-11  were  built 
so  as  to  take  either  tension  or  compression,  the  counter 
9-12  would  not  be  needed.  In  this  latter  case  the  member 
10-11  would  have  a  range  of  stress  from  —  21.85  tons  to 
+  9.87  tons,  the  former  when  the  live  load  covered  all  the 
joints  on  the  right  of  it,  the  latter  when  it  covered  all  the 
joints  on  the  left.  But  as  the  diagonals  are  to  be  subjected 
only  to  compression  in  this  truss,  the  minimum  stress  in 
10-11  is  zero,  and  the  tension  of  9.87  tons  in  10-11  becomes 
the  maximum  compression  in  the  counter  brace  9-12.  The 
three  diagonals,  4-5,  6-7,  8-9,  are  always  in  compression. 


BRIDGE  TRUSSES.  77 

Hence,  theoretically,  there  is  only  one  counter  brace  needed 
in  the  left  half  of  this  truss;  but,  practically,  a  counter 
brace  would  be  put  in  the  fourth  panel  also,  so  that  the 
truss  would  have  counter  braces  in  its  four  middle  panels. 

The  maximum  stress  in  any  vertical,  except  the  middle 
one,  is  equal  to  the  maximum  positive  shear  in  the  section 
cutting  it  and  two  chord  members,  or  it  is  equal  to  the 
maximum  positive  shear  in  the  panel  toward  the  abutment 
from  the  vertical.  Thus, 

Max.  stress  in  3-4  =  13  x  4.5  =  58.5  tons. 
Max.  stress  in  5-6 

=  [4  x  3.5  +  .9  (1  +  2  +  ...  +  8)]  =  46.4  tons. 
.Max.  stress  in  7-8 

=  [4  x  2.5  +  -9  (1  +  2  +  -.  +  7)]  =  35.2  tons. 
Max.  stress  in  9-10 

=  [4  x  1.5  +  .9(1  +2  +  ...  +  6)]  =  24.9  tons. 

The  maximum  stress  in  the  middle  vertical  11-12  is  equal, 
either  to  the  maximum  positive  shear  in  panel  10-12,  or  to 
a  full  panel  dead  and  live  load,  whichever  is  the  greater. 
For,  if  this  maximum  positive  shear  is  greater  than  a  panel 
load,  then  the  shear  in  the  next  right  hand  panel  under  the 
same  loading  is  also  positive,  and  the  counter  brace  in  that 
panel  will  be  in  action,  thus  making  the  shear  in  panel  10-12 
the  same  as  that  in  the  vertical  11-12 ;  but,  when  the  coun- 
ters are  not  in  action,  the  stress  in  11-12  is  always  equal  to 
the  load  at  12. 

Max.  positive  shear  in  10-12 

=  4  x  0.5  +  ^(1  +  2  +  ...  +  5)  =  15.5  tons. 
Full  panel  load  =  4  +  9  =  13  tons. 
.-.  Max.  stress  in  11-12  =  15.5  tons. 


78 


ROOFS  AND   BRIDGES. 


The  minimum  stress  in  any  vertical  is  equal  to  the  maxi- 
mum negative  shear  in  the  section  cutting  it  and  two  chord 
members,  or  it  is  equal  to  the  maximum  negative  shear  in 
the  panel  toward  the  abutment  from  the  vertical,  or  to  a 
dead  panel  load,  whichever  is  the  greater.  Thus, 

Min.  stress  in  3-4    =  4  x  4.5  -  0  =  18.0  tons. 
Min.  stress  in  5-6    =  4  x  3.5  —  -^  x  1  =  13.1  tons. 
Min.  stress  in  7-8    =  4  x  2.5  -  ^  (1  +  2)  =  7.3  tons. 
Min.  stress  in  9-10  =  4  x  1.5  —  •&(!+  2  +  3)  =  0.6  tons. 

But  the  stress  in  a  vertical  of  this  truss  cannot  be  less 
than  the  weight  of  a  dead  panel  load ;  therefore 

Min.  stress  in  9-10  =  4  tons. 
Min.  stress  in  11-12  =  4  tons. 

We  may  now  collect  these  web  stresses  in  a  table  as 
follows : 

WEB  STRESSES. 


MEMBERS. 

2-3 

4-5 

6-7 

8-9 

10-11 

9-12 

Max.  stresses 
Min.  stresses 

-82.48 
-25.38 

-65.42 
-18.47 

-49.63 
-10.29 

-35.10 
-  0.85 

-21.85 
0.00 

-9.87 
0.00 

VERTICALS. 


MBMBEES. 

3-4 

5-8 

7-8 

9-10 

11-12 

Max.  stresses  .... 

58.5 

46.4 

35.2 

24.9 

15.5 

Min.  stresses    .... 

18.0 

13.1 

7.3 

4.0 

4.0 

BRIDGE   TRUSSES.  79 

In  building  Howe  trusses  of  timber,  it  is  usually  the 
practice  to  put  in  all  the  dotted  diagonals  to  keep  the  truss 
rigid.  This  is  not  done  in  metallic  trusses. 

Prob.  60.   A  through  Howe  truss  has  11  panels,  each  10 
feet  long  and  10  feet  deep ;  the  dead  load  is  800  Ibs.,  and 
the  live  load  is  1600  Ibs.  per  foot  per  truss :  find  the  maxi- 
mum and  minimum  stresses  in  all  the  members. 
Ans.  Max.  stresses  in  lower  chord 

=  60,  108,  144,  168,  180,  and  180  tons. 
Min.  stresses  in  lower  chord 

=  20,  36,  48,  56,  60,  and  60  tons. 
Max.  stresses  in  main  braces 

=  -84.60,  -68.71,  -53.83,    -40.0,   -27.18, 

- 15.37  tons. 
Min.  stresses  in  main  braces 

=  _28.20,  -21.53,  -13.85,  -5.13,  -0.00,  -0.00  tons. 
Max.  stresses  in  counters  =  —  4.36,  —  15.38  tons. 

Max.  stresses  in  verticals 

=  +  60.0,  +48.73,  +38.18,  +28.36,  +19.28  tons. 
Min.  stresses  in  verticals 

=  +  20.0,  +  15.27,  +  9.82,  +  4.00,  +  4.00  tons. 

Prob.  61.   A  deck  Howe  truss  has  12  panels,  each  10  feet 
long  and  10  feet  deep ;  the  dead  load  is  600  Ibs.,  and  the 
live  load  is  1200  Ibs.  per  foot  per  truss:    find  the  maxi- 
mum and  minimum  stresses  in  all  the  members. 
Ans.  Max.  stresses  in  lower  chord 

=  49.5,  90.0,  121.5, 144.0, 157.5,  162  tons. 
Max.  stresses  in  main  braces 

=  -69.79,  -57.81,  -46.53,  -35.95,  -26.08,  -16.9  tons. 


80  ROOFS  AND   BRIDGES. 

Min.  stresses  in  main  braces 

=  -  23.25,  -  18.33,  - 12.69,  -  6.35,  0.0,  0.0  tons. 

Max.  stresses  in  counters  =  —  0.71,  —  8.46  tons. 
Max.  stresses  in  verticals 

=  4.5,  41.0,  33.0,  25.5, 18.5, 12.0,  6.0  tons. 

The  maximum  stress  in  any  vertical  of  the -deck  Howe 
truss  except  the  middle  one,  is  equal  to  the  maximum  posi- 
tive shear  in  the  section  cutting  it  and  two  chord  members, 
or  it  is  equal  to  the  maximum  positive  shear  in  the  panel 
away  from  the  abutment  from  the  vertical.  The  stresses  in 
the  verticals  need  to  be  very  carefully  tested. 

Prob.  62.  A  deck  Howe  truss  has  10  panels,  each  14  feet 
long  and  14  feet  deep;  the  dead  load  is  714  Ibs.  and  the 
live  load  is  2000  Ibs.  per  foot  per  truss :  find  the  maximum 
and  minimum  stresses  in  all  the  members. 

Ans.  Max.  stresses  in  lower  chord 

=  85.5,  152.0,  199.5,  228.0,  237.5  tons. 
Max.  in  main  braces 

=  -  120.55,  -  95.74,  -  72.89,  -  52.02,  -  33.14  tons. 
Min.  in  main  braces 

=  -  31.73,  -  22.70,  -  11.70,  0.0,  0.0  tons. 
Max.  in  verticals 

=  9.5,  67.9,  51.7,  36.9,  23.5, 11.5  tons. 
Max.  in  counters  =  —  1.27,  —  16.22  tons. 

Prob.  63.  A  through  Howe  truss  has  8  panels,  each  18 
feet  long  and  24  feet  deep ;  the  dead  load  is  806  Ibs.,  and 
the  live  load  is  1500  Ibs.  per  foot  per  truss :  find  the 
maximum  and  minimum  stresses  in  all  the  members. 


BRIDGE   TRUSSES.  81 

Art.  26.  The  Pratt  Truss.  — All  parts  of  the  Pratt 
truss  are  of  iron  or  steel,  the  verticals  being  compression 
members,  except  the  hip  verticals,  and  the  diagonals  ten- 
sion members.  The  maximum  and  minimum  stresses  of 
any  member  may  be  determined  directly  from  a  single  equa- 
tion, as  in  the  case  of  the  Howe  truss  (Art.  25). 

Prob.  64.  A  through  Pratt  truss,  as  a  through  railroad 
bridge,  Fig.  33,  has  10  panels,  each  14  feet  long  and  14  feet 

i t- 


10         12        10' 
Fig.  33 

deep ;  the  dead  load  from  formula  (1),  Art.  15,  the  live  load 
is  2300  Ibs.  per  foot  per  truss:  find  the  maximum  and 
minimum  stresses  in  all  'the  members. 

From  formula  (1),  Art.  15,  the  dead  panel  load  per  truss 

=  (9x140  +  520)14  =  12>460  lbs_  =  6  23  tons  =  say)  6  tong> 

r 
Live  panel  load  per  truss  =  16.1  tons  =  say,  16  tons. 

tan  6  =  1,  sec  0  =  1.414. 

The  max.  and  min.  chord  stresses  may  be  written  out  at 
once  (Arts.  19  and  21).     They  are  the  following : 

Max.  stresses  in  lower  chord 

=  99.0,  99.0,  176.0,  231.0,  264.0  tons. 
Min.  stresses  in  lower  chord 

=  27.0,  27.0,  48.0,  63.0,  72.0  tons. 


82 


ROOFS  AND  BRIDGES. 


WEB  STRESSES. 

The  web  stresses  may  be  found  exactly  as  in  Prob.  60. 
The  following  are  the  equations  for  a  few  of  the  members : 

Max.  stress  in  2-3  =  -  [4.5  x  22]1.41  =  -  139.5  tons. 
Max.  stress  in  3-6 

=  [3.5  x  6  +  1.6(1  +  2  +  ...  +  8)]1.41  =  110.8  tons. 
Max.  stress  in  5-8 

=  [2.5  x  6  +  1.6(1  +  2  +  —  +  7)]1.41  =  84.3  tons. 
Max.  stress  in  7-8 

=  -[1.5  x  6  + 1.6(1  +•  2  +  .-  +•  6)]  =  -  42.6  tons. 
Max.  stress  in  11-12 

=  -  [-0.5  x  6  +  1-6(1  +  2  +  ...+•  4)]=  - 13.0  tons. 
Min.  stress  in  5-8 

=  [2£  x  6  -  1.6(1  +  2)]1.41  =  14.4  tons. 
Max.  stress  in  10-11 

=  [_  i  x  6  + 1.6(1  +  2  +  ...  +  4)]1.41  =  18.5  tons. 

Thus  we  find  the  stresses  in  the  following  table : 
WEB  STRESSES. 


MEMBERS. 

2-3 

3-6 

5-S 

7-10 

9-12 

8-9 

10-11 

Max.  stresses 
Min.  stresses 

-139.5 

-  38.1 

+  110.8 
+  27.4 

+84.3 

+  14.4 

+60.0 
0.0 

+38.0 
0.0 

+0.8 
0.0 

+18.5 
0.0 

Max.  stresses  in  the  verticals  =  +•  22.0,  —  59.8,  —  42.6, 
—  27.0,  -13.0  tons. 


BRIDGE   TRUSSES.  83 

Prob.  65.  A  through  Pratt  truss  has  11  panels,  each  10 
feet  long  and  10  feet  deep ;  the  dead  load  is  1200  Ibs.,  and 
the  live  load  is  2000  Ibs.  per  foot  per  truss :  find  the 
stresses  in  all  the  members. 

Ans.  Max.  stresses  in  lower  chord 

=  80,  80,  144,  192,  224,  240  tons. 
Max.  in  main  diagonals 

=  -  113.1,  +  91.5,  +  71.5,  +  52.8,  +  35.4,  + 19.2  tons. 
Min.  in  main  diagonals 

=  _42.3,  +32.6,  +21.5,  +9.2,  0.0,  0.0  tons. 
Max.  in  counters  =  +  4.4,  +  19.21  tons. 
Max.  in  verticals 

=  _|_  16.0,  -  50.7,  -  37.4,  -  25.1,  -  13.6  tons. 

Prob.  66.  A  deck  Pratt  truss,  Fig.  34,  has  10  panels, 
each  12  feet  long  and  12  feet  deep ;  the  dead  load  is  833 


Ibs.,  and  the  live  load  is  1667  Ibs.  per  foot  per  truss :  find 
the  stresses  in  all  the  members. 

Ans.  Max.  stresses  in  lower  chords 

=  0.0,  67.5,  120.0,  157.5,  180.0  tons. 
Max.  in  main  ties  =  95.1,  75.4,  57.1,  40.2,  24.6  tons. 
Min.  in  main  ties  =  31.7,  23.2,  13.4,  2.1,  0.0  tons. 
Max.  in  counters  =  10.6  tons. 
Max.  in  verticals 

=  _  75.0,  -  67.5,  -  53.5,  -  40.5,  -  28.5,  -  17.5  tons. 


84  ROOFS  AND  BRIDGES. 

Prob.  67.  A  deck  Pratt  trass  has  9  panels,  each  18  feet 
long  and  24  feet  deep ;  the  dead  load  is  500  Ibs.,  and  the 
live  load  is  1000  Ibs.  per  foot  per  truss :  find  the  stresses 
in  all  the  members. 

-4ns.  Max.  stresses  in  lower  chords 

=  0.0,  40.5,  70.91,  91.1,  101.3  tons. 
Max.  stresses  in  main  ties 

=  67.5,  51.9,  37.5,  24.4,  12.5  tons. 
Min.  stresses  in  main  ties 

=  22.5,  15.6,  7.5,  0.0,  0.0  tons. 
Max.  stresses  in  counters  =  1.9,  12.5  tons. 
Max.  stresses  in  verticals 

=  _  60.8,  -  54.0,  -  41.5,  -  30.0,  -  19.5  tons. 

The  chord  stresses  are  the  same  in  the  Howe  trusses,  and 
also  in  the  Pratt  trusses,  whether  the  load  be  placed  on  the 
top  or  the  bottom  chord. 

Art.  27.  The  Warren  Truss  with  Vertical  Sus- 
penders.—  The  left  half  of  this  truss  is  represented  in 
Fig.  35,  in  which  the  web  bracing  consists  of  equilateral  or 


isosceles  triangles  and  vertical  ties  or  suspenders.  In  a 
through  truss,  each  of  the  verticals  3,  5,  7,  9,  simply  carries 
a  panel  load  from  the  lower  chord  to  the  upper,  while  the 
verticals  4,  6,  8,  10,  are  only  to  support  and  stiffen  the 
upper  chord.  If  the  truss  is  a  deck,  the  last  named  ver- 
ticals become  posts. 


BRIDGE   TRUSSES.  85 

Prob.  68.  A  through  Warren  truss  with  vertical  ties, 
one  half  of  which  is  shown  in  Fig.  35,  has  16  panels,  each 
8  feet  long,  its  braces  all  forming  equilateral  triangles  ;  the 
dead  load  is  given  by  formula  (2)  of  Art.  15,  the  live  load 
is  1500  Ibs.  per  foot  per  truss  :  find  the  maximum  and 
minimum  stresses  in  all  the  members. 

The  dead  panel  load  per  truss 


Live  panel  load  per  truss  =  150°  x  8  =  6  tons. 
2000 

tan  0  =  .577  ;  sec  0  =  1.1547. 

The  chord  stresses  may  be  written  out  at  once  by  chord 
increments. 

For  the  verticals  the  maximum  stress  is  evidently  the 
dead  and  live  panel  load,  or  9  tons;  the  minimum  stress 
is  the  dead  panel  load,  or  3  tons. 

The  following  are  the  equations  for  determining  the 
stresses  in  a  few  of  the  members  : 

Max.  stress  in  2-4  =  7.50  x  9  x  .577  =  38.9  tons. 
Max.  stress  in  4-6 

=  [7.50  +  6.50  +  5.50]9  x  .577  =  101.3  tons. 
Max.  stress  in  3-5 

=  -[7.50  +  6.50]  x  9  x  .577=  -  72.7  tons. 
Max.  stress  in  2-3  =  -[7.50  x  9  x  1.154]=  -  77.9  tons. 
Max.  stress  in  5-6 

=  [4.5  x  3  +  ^(1  +  2  +  3  +  -4-12)]  1.154 
=  42.75  x  1.154  =  +  49.3  tons. 
Min.  stress  in  7-8 


In  this  way  all  the  stresses  may  be  found. 


86 


ROOFS  AND  BRIDGES. 

CHOKD  STRESSES. 


MEMBERS. 

8-5 

5-7 

7-9 

9-9 

2-4 

4-6 

6-8 

8-10 

Max.  stresses 

-72.7 

-124.6 

-155.8 

-166.2 

+  88.9 

+  101.8 

+  142.8 

+  163.6 

Min.  stresses 

-24.2 

-  41.5 

-  51.9 

-  55.4 

+  13.0 

+   33.8 

+   47.6 

+   54.5 

WEB  STRESSES. 


MEMBERS. 

2-8 

3-4 

4-5 

5-6 

C-7 

7-8 

8-9 

9-10 

Max.  stresses  .... 

-77.9 

+  67.9 

-58.4 

+  49.3 

-40.6 

+  32.5 

-24.7 

+  17.3 

Min.  stresses  .... 

-26.0 

+  22.1 

-17.7 

+  18.0 

-  7.8 

+   2.2 

+  3.9 

-10.4 

We  see  from  this  table  of  web  stresses  that  the  members 
8-9  and  9-10  should  be  capable  of  resisting  both  tension  and 
compression,  and  hence  should  be  counter  braces  (Art.  1) ; 
and,  of  course,  the  same  is  true  for  the  right  half  of  the  truss. 
Prob.  69.  A  through  Warren  truss  with  vertical  ties,  like 
Fig.  35,  has  20  panels,  each  10  feet  long,  its  braces  all 
forming  equilateral  triangles;  the  dead  load  is  given  by 
formula  (2)  of  Art.  15,  the  live  load  is  1600  Ibs.  per  foot 
per  truss :  find  the  maximum  and  minimum  stresses  in  all 
the  members. 

Ans.  Max.  stresses  in  upper  chord 

=  - 135.0,  -  240.0,  -  315.0,  -  360.0,  -  375.0  tons. 
Max.  stresses  in  lower  chord 

=  +  71.3,  +191.3,  +281.3,  +341.3,  +371.3  tons. 
Max.  web  stresses 

= -142.5,  +128.0,  -113.9,  +100.2,  -87.1,  +74.4, 

-  62.2,  +  50.4,  -  39.1,  +  28.3  tons. 
Min.  web  stresses 

=  -54.8,  +48.6,  -41.9,  +34.7,  -27.1,  +19.0,  -10.5, 
+  1.5,  +8.0,  -17.9  tons. 


BRIDGE   TRUSSES.  87 

Prob.  70.  A  through  Warren  truss  with  vertical  ties, 
like  Fig.  35,  has  10  panels,  each  12  feet  long  and  12 
feet  deep,  its  braces  all  forming  isosceles  triangles ;  the 
dead  and  live  loads  are  1000  Ibs.,  and  2000  Ibs.  per  foot  per 
truss :  find  the  maximum  and  minimum  stresses  in  all  the 
members. 

Art.  28.  The  Double  Warren  Truss,  or  Double  Tri- 
angular Truss,  or  Girder.  —  This  truss,  shown  in  Fig.  36, 
has  two  systems  of  triangular  bracing,  the  one  system 
represented  in  full  lines  and  the  other  in  dotted  lines,  the 


j    A     -    J\    °    *1    ?'   I'    £'    1'    V 


chords  being  common  to  both  systems.     The  roadway  may 
be  carried  either  upon  the  upper  or  the  lower  chord. 

In  all  the  trusses  hitherto  considered,  a  vertical  section 
taken  at  any  point  of  the  truss  will  cut  only  three  members ; 
but  in  the  double  Warren  truss  this  is  not  the  case,  since  a 
vertical  section  will  in  general  cut  four  members,  causing 
the  problem  at  first  to  seem  to  be  indeterminate  (Art.  7). 
This  is  obviated,  however,  by  the  fourth  condition  that: 
the  truss  is  equivalent  to  the  two  trusses,  shown  in  full  and 
in  dotted  lines,  welded  into  one.  Thus,  the  loads  at  6,  10, 
10',  6'  are  carried  to  the  abutments  by  the  diagonals  drawn 
in  full,  while  the  loads  at  4,  8,  12,  8',  4r  are  carried  by  the 
dotted  diagonals.  The  chord  and  web  stresses  are  readily 
found  as  in  a  simple  triangular  truss,  assuming  each  system 
as  independent.  This  truss  is  used  both  as  a  riveted,  and 
as  a  pin-connected  bridge. 


88 


ROOFS  AND  BRIDGES. 


Prob.  71.  A  double  Warren  truss,  Fig.  36,  as  a  deck  rail- 
road bridge,  has  10  panels,  each  14  feet  long  and  14  feet 
deep ;  the  dead  load  is  given  by  formula  (2)  of  Art.  15,  the 
live  load  is  2000  Ibs.  per  foot  per  truss :  find  the  stresses  in 
all  the  members. 

The  dead  panel  load  per  truss 

A"|±««0^  14  =  11,060  Ibs.  =5.5  tons. 

Live  panel  load  per  truss  =  14  tons. 

tan  0  =  1,  sec  0  =  1.414. 

The  following  are  the  equations  for  determining  the 
stresses  in  a. few  of  the  members: 

Max.  stress  in  1-3  =  —  2  x  19.5  x  1  =  —  39  tons. 
Max.  stress  in  3-5  =  —  [2.0  +•  2.5  +•  1.5]  19.5  =  — 117  tons. 
Max.  stress  in  6-8  =  [2.5  +•  4.0  +  3.0]  19.5  =  + 185.2  tons. 
Max.  stress  in  3-6 

=  [1.5  x  5.5  +•  1.4  (1  +  3  +5  +-  7)]  1.414=  +  43.3  tons. 
Min.  stress  in  5-8  =  [5.5  - 1.4  x  2]  1.414  =  +  3.8  tons. 
Thus  the  following  stresses  are  determined : 

UPPER  CHORD  STRESSES. 


MEMBERS. 

1-3 

3-5 

5-7 

7-9 

9-11 

Max.  stresses    .  .  . 

-39.0 

-117.0 

-175.5 

-214.5 

-234.0 

Min.  stresses    .  .  . 

-11.0 

-  33.0 

-  49.5 

-  60.5 

-  66.0 

LOWER  CHORD  STRESSES. 


MEMBERS. 

2-4 

4-6 

C-8 

8-10 

10-12 

Max.  stresses    .  .  . 
Min.  stresses    .  .  . 

+48.8 
+  13.8 

+  126.8 
+  35.8 

+  185.2 
+  52.3 

+224.3 
+  63.3 

+243.8 
+  68.8 

BRIDGE   TRUSSES.  89 

WEB  STRESSES  (DOTTED  SYSTEM). 


MEMBERS. 

1-4 

4-6 

5-8 

8-9 

9-12 

Max  stresses    ... 

+55.2 

-55.2 

+31  5 

-31  5 

+  11  9 

Min.  stresses  

+  15.6 

-15.6 

+  3.8 

O  Q 
—    O.O 

-11.9 

WEB  STRESSES  (FULL  SYSTEM). 


MEMBERS. 

2-3 

3-6 

6-7 

7-10 

10-11 

Max  stresses  .  . 

-69.0 

+  43  3 

—433 

+  21  7 

21  7 

Min.  stresses  

-19.4 

+  9.7 

-  9.7 

-   4.1 

+  4.1 

Prob.  72.  A  double  deck  Warren  truss,  like  Fig.  36,  has 
12  panels,  each  10  feet  long  and  10  feet  deep ;  the  dead  load 
is  800  Ibs.,  and  the  live  load  is  1600  Ibs.  per  foot  per  truss : 
find  the  stresses  in  all  the  members. 

Am.  Max.  stresses  in  upper  chord 

=  -.30,  -90,  -138,  -174,  -198,  -210  tons. 

Max.  stresses  in  lower  chord 

=  36,  96,  144, 180,  204,  216  tons. 

Max.  stresses  in  dotted  diagonals 

=  +  42.4,  -  42.4,  +  27.3,  -  27.3,  +  14.1,  -  14.1  tons. 

Max.  stresses  in  full  diagonals 

=  _50.9,  +34.9,  -34.9,  +20.7,  -20.7,  +8.5  tons. 

Min.  stresses  in  dotted  diagonals 

=  +  14.1,  -14.1,  +6.6,  -6.6,  -2.8,  +  2.8tons. 

Min.  stresses  in  full  diagonals 

=  _  17.0,  + 10.4,  - 10.4,  +  1.9,  -  1.9,  -  8.5  tons. 


90 


ROOFS  AND  BRIDGES. 


Prob.  73.   A  double  through  Warren  truss,  like  Fig.  36, 
has  10  panels,  each  10  feet  long  and  10  feet  deep ;  the  dead 
and  live  loads  are  800  Ibs.,  and  2000  Ibs.  per  foot  per  truss : 
find  the  stresses  in  all  the  members. 
Ans.  Max.  stresses  in  upper  chord 

=  _  35,  _  91,  _  133,  - 161,  - 175  tons. 
Max.  stresses  in  lower  chord 

=  +  28,  +  84,  +  126,  + 154,  + 168  tons. 
Max.  stresses  in  diagonal  ties 

=  +  49.4,  +39.6,  +31.1,  +22.5,  +15.6,  +  8.5  tons. 
Min.    stresses  in  diagonal  ties 

=  +  14.1,   +11.3,   +7.1,  +2.8,  -2.9,  -8.4  tons. 

Prob.  74.  A  double  through  Warren  truss,  like  Fig.  36, 
has  12  panels,  each  15  feet  long  and  15  feet  deep ;  the  dead 
load  is  given  by  formula  (3)  of  Art.  15,  the  live  load  is 
2000  Ibs.  per  foot  per  truss  :  find  the  stresses  in  all  the 
members. 

Art.  29.  The  Whipple  Truss.  —  This  truss,  shown  in 
Fig.  37,  consists  of  two  simple  Pratt  trusses  combined. 


The  ties  in  the  web  system  extend  over  two  panels,  and  it 
is  therefore  often  called  a  "  double  intersection  Pratt  truss."* 
The  advantage  over  the  Pratt  for  long  spans  is  that  it  has 
short  panels,  while  keeping  the  inclination  of  the  diagonals 
at  about,  45°. 

*  Called  also  the  Linville  Truss. 


BRIDGE   TRUSSES.  91 

This  truss  was  at  one  time  more  common  than  any  other 
in  American  bridge  practice.  It  is  still  used  though  very 
rarely  for  highway  bridges;  for  railway  bridges  it  is  as  a 
rule  avoided  by  the  best  practice. 

It  is  assumed  that  this  truss  is  equivalent  to  the  two 
trusses  shown  in  full  and  in  dotted  lines,  and  that  each 
system  acts  independently  and  carries  only  its  own  loads  to 
the  abutments ;  but  in  the  case  of  the  web  members,  this  is 
not  strictly  true.  If,  however,  the  truss  is  built'with  an  even 
number  of  panels,  the  error  in  the  web  stresses  obtained  on 
the  assumption  of  independent  systems  is  probably  very 
small.  With  the  chord  stresses  there  is  no  ambiguity ;  the 
chord  stresses  are  a  maximum  for  a  full  load,  and  may  be 
found  as  usual  by  chord  increments,  or  by  moments.  The 
vertical  shear  will  be  the  same  whether  the  load  is  applied 
at  the  top  or  at  the  bottom  chord. 

Prob.  75.  A  through  Whipple  truss,  Fig.  37,  has  12  panels, 
each  10  feet  long  and  20  feet  deep ;  the  dead  load  is  750  Ibs., 
and  the  live  load  is  1200  Ibs.  per  foot  per  truss :  find  the 
stresses  in  all  the  members. 

The  dead  panel  load  per  truss  =  3.75  tons. 

The  live  panel  load  per  truss   =  6.0  tons. 

tan0  =  l:  sec 0  =  1.414:  tan 0' =  0.5:  sec $' =  1.118. 

The  following  are  the  equations  for  finding  the  stresses  in 
a  few  of  the  members : 

Max.  stress  in  4-6  =  3  x  9.75  x  0.5  =  14.6  tons. 
Max.  stress  in  6-8  =  14.6  +  2£  x  9.75  x  1  =  39.0  tons 

=  —  stress  in  1-3. 
Max.  stress  in  9-11 

=  -  [39.0  +  (2  + 1£  + 1  +  £)  9.75]=  -  87.8  tons 

=  stress  in  11-13, 


92 


ROOFS  AND  BRIDGES. 


since  for  a  uniform  load  the  diagonals  meeting  the  upper 
chord  between  9  and  9'  are  not  in  action. 

Max.  stress  in  1-4  =  3  x  9.75  x  1.118  =  32.7  tons. 
Max.  stress  in  1-6 

=  [2£  x  3.75  +  ^(2  +  4  +  6  +  8  + 10)]  x  1.414 

=  24.375  x  1.414  =  34.5  tons. 

Max.  stress  in  1-2  =  -  29.25  -  24.375  =  -  53.6  tons. 
Thus  the  following  stresses  are  determined : 
CHORD  STRESSES. 


MEMBERS. 

2-1 

4-6 

6-8 

8-10 

10-12 

12-14 

9-13 

Max.      . 

0.00 

14.6 

39.0 

58.5 

73.1 

82.9 

87.8 

Min.       . 

0.00 

5.6 

15.0 

22.6 

28.1 

31.8 

33.8 

STRESSES  IN  DIAGONALS. 


MEMBERS. 

1-4 

1-6 

3-8 

5-10 

7-12 

9-14 

11-12' 

18-10' 

Max.       . 

32.7 

34.5 

28.3 

22.1 

16.6 

11.1 

6.4 

1.6 

Min. 

12.6 

13.3 

9.9 

6.5 

2.5 

0.0 

0.0 

0.0 

STRESSES  IN  POSTS. 


MEMBERS. 

1-2 

3-1 

5-6 

7-8 

9-10 

11-12 

18-14 

Max.      . 

53.6 

20.0 

15.6 

11.8 

7.9 

4.5 

1.1 

Min.       . 

20.6 

7.0 

4.6 

1.8 

0.0 

0.0 

0.0 

Prob.  76.  A  through  Whipple  truss,  Fig.  38,  has  16 
panels,  each  10  feet  long  and  20  feet  deep ;  the  dead  and 
live  loads  are  800  Ibs.,  and  1600  Ibs.  per  foot  per  truss: 
find  the  stresses  in  all  the  members. 


BRIDGE  TRUSSES. 


93 


Ans. 


8        10       12       Ik       16       18       18' 
Fig.  38 

CHORD  STRESSES. 


MEMBERS. 

4-6 

6-8 

S-10 

10-12 

12-14 

14-16 

16-18 

13-17 

Max.        . 

24. 

66. 

102. 

132. 

156. 

174. 

186. 

192. 

Min.  .     . 

8. 

22. 

34. 

44. 

52. 

58. 

62. 

64. 

STRESSES  IN  FULL  DIAGONALS. 


MEMBERS. 

1-6 

5-10 

0-14 

13-18 

17-14' 

Max.     .    ,    . 

59.4 

43.8 

29.7 

17.0 

5.7 

Min.      .     .     . 

19.8 

.12.7 

4.2 

0.0 

0.0 

STRESSES  IN  DOTTED  DIAGONALS. 


MEMBERS. 

1-4 

3-8 

7-12 

11-16 

15-16' 

15-12 

Max.    .     . 

63.7 

51.6 

36.8 

23.3 

11.3 

0.7 

Min.    .    . 

17.9 

16.3 

8.5 

0.0 

0.0 

0.0 

STRESSES  IN  POSTS. 


MEMBERS. 

M 

3-4 

6-6 

7-8 

9-10 

11-12 

18-14 

15-16 

17-18 

Max.      .     . 

90.0 

36.6 

31.0 

26.0 

21.0 

16.6 

12.0 

8.0 

4.0 

Min.       .     . 

30.0 

11.5 

9.0 

6.0 

3.0 

0.0 

0.0 

0.0 

0.0 

94  ROOFS  AND   BRIDGES. 

Prob.  77..  A  deck  Whipple  truss,  like  Fig.  38,  has  16 
panels,  each  12  feet  long  and  24  feet  deep;  the  dead 
load  is  750  Ibs.  per  foot  per  truss,  and  the  live  load  is 
2000  Ibs.  per  foot  per  truss:  find  the  stresses  in  all  the 
ir.embers. 

Art.  30.  The  Lattice  Truss,  or  Quadruple  Warren 
Truss.*  —  This  truss,  Fig.  39,  contains  four  web  systems 


LAy\y\/Xy'X/X//X/X/'X/xxX/Ay\y\y\yX/'X/\y\xX 

2  ^    It     6     8     10   la    IU   16   18  20   22   20'  18'  16'  lit'   1*'  10'  3'    6'  4'        4 


welded  into  one.     It  is  built  only  as  a  short-span  riveted 
structure. 

In  determining  the  stresses  in  the  different  members  of 
this  truss,  an  ambiguity  arises  from  two  causes :  (1)  the 
web  members  being  riveted  together  at  each  intersection, 
the  different  systems  cannot  act  independently ;  and  (2)  two 
of  the  systems  are  not  symmetrically  placed  in  reference  to 
the  center  of  the  truss.  Thus,  if  equal  loads  be  placed  at 
each  joint,  the  abutment  at  2  will  carry  more  than  half  of 
the  load  on  the  system  4,  12,  20,  16',  8',  and  less  than  half 
of  the  load  on  the  system  8, 16,  20',  12',  4'.  But  it  simplifies 
the  solution  to  assume  that  each  system  acts  independently, 
and  is  symmetrical  in  reference  to  the  center  of  the  truss. 
The  chord  and  web  stresses  are  then  readily  found,  as  in  a 
simple  triangular  truss. 

*  All  double  systems,  such  as  this  and  the  Whipple,  owing  to  the  inde- 
terminate character  of  the  strains,  are  now  usually  avoided  by  the  best 
practice. 


BEIDGE  TRUSSES.  95 

Prob.  78.  A  through  lattice  truss  containing  four  web 
systems,  as  shown  in  Fig.  39,  has  20  panels,  each  10  feet 
long,  the  depth  of  the  truss  is  20  feet;  the  dead  load  is 
given  by  formula  (2)  Art.  15,  the  live  load  is  2000  Ibs.  per 
foot  per  truss  :  find  the  stresses  in  all  the  members. 

Dead  panel  load  per  truss  =  5  tons. 

Live  panel  load  per  truss  =  10  tons. 

tan  0  =  1,  sec  0  = 


NOTE.  —  In  finding  the  shear  for  a  web  member  due  to  the  dead 
load  or  to  the  live  load  covering  the  whole  truss,  a  slight  difficulty 
arises,  owing  to  the  two  unsymmetrical  systems  above  referred  to. 
This  shear  may  be  found,  either  by  taking  the  algebraic  sum  of  the 
greatest  positive  and  the  greatest  negative  shears  (Art.  22),  or  by 
simple  inspection  of  the  truss. 

Thus,  the  dead  load  shear  in  5-10  is  the  weight  of  the  two  panel 
loads  at  the  joints  10  and  18,  because  the  system  to  which  5-10  be- 
longs is  symmetrical  with  reference  to  the  center  of  the  truss  ;  but 
the  shear  in  7-12  is  not  the  weight  of  the  two  panel  loads  at  the  joints 
12  and  20,  because  this  system  is  not  symmetrical  in  reference  to  the 
center.  If  the  joint  20  were  at  the  center  22,  the  shear  for  7-12  would 
be  \\  panel  loads  ;  but  if  it  were  at  the  joint  18,  the  shear  would  be 
2  panel  loads.  Being  midway  between  the  joints  18  and  22,  the  shear 
for  7-12  is  the  mean  of  \\  and  2  panel  loads,  or  1.75  panel  loads  ;  and 
so  for  the  shear  in  any  other  web  member. 

By  chord  increments,  we  have 

Max.  stress  in  2-4  =  2  x  15  x  1  =  30  tons. 

Max.  stress  in  4-6 

=  (2  +  1-75  +  2.75)  x  15  =  97.5  tons. 
Max.  stress  in  6-8 

=  97.5  -(-  (1.5  +  2.5)  x  15  =  157.5  tons. 
Max.  stress  in  1-3  =  2.5  x  15  =  37.5  tons. 
Max.  stress  in  3-5  =  (2.5  +  2  x  2.25)  x  15  =  105.0  tons. 


96 


BOOFS  AND  BRIDGES. 


MAXIMUM  WEB  STRESSES. 

The  maximum  stress  in  3-8  will  be  when  the  dead  and 
live  loads  cover  the  whole  system  to  which  3-8  belongs. 
Likewise  for  the  member  5-10.  The  maximum  live  load 
stress  in  7-12  will  be  when  the  live  panel  loads  are  at  the 
joints  12,  20,  16',  8',  since  these  are  the  only  loads  which 
act  in  the  system  to  which  7-12  belongs,  on  the  right  of 
7-12.  For  the  dead  load  stress,  see  Note. 

Max.  stress  in  3-8 

=  2.25  x  15  x  1.414  =  47.7  tons  =  -  stress  in  A-3. 
Max.  stress  in  5-10 

=  2  x  15  x  1.414  =  42.4  tons  =  -  stress  in  2-5. 
Max.  stress  in  7-12 

=  [1.75x5+ ^£(3 +7 +11 +15)]  1.414  =  37.8  tons 

=  —  stress  in  4-7. 
Max.  stress  in  9-14 

=  [1.5  x5  +  £(2  +  6  +  10  + 14)]  1.414  =  33.2  tons 

=  —  stress  in  6-9. 

Thus  the  following  stresses  are  determined : 
UPPER  CHORD  STRESSES. 


MEMBERS. 

1-3 

3-5 

5-7 

7-9 

9-11 

Max.     .     .     . 
Min.     .     .     . 

37.5 
12.5 

105.0 
35.0 

165.0 
55.0 

217.5 
72.5 

262.5 

87.5 

MEMBERS. 

11-13 

13-15 

15-17 

17-19 

19-21 

Max.     .    .    . 
Min.     .     .    . 

300.0 
100.0 

330.0 
110.0 

352.5 
117.5 

367.5 
122.5 

375.0 
125.0 

BKIDGE  TEUSSES. 
LOWER  CHORD  STRESSES. 


97 


MEMBERS. 

8-4 

4-0 

6-8 

S-10 

10-12 

Max.     .    .    . 

Min.      .     .     . 

30.0 
10.0 

97.5 
32.5 

157.5 
52.5 

210.0 
70.0 

255.0 
85.0 

MEMBERS. 

12-14 

14-16 

16-18 

18-20 

£0  -'22 

Max.     .     .   :. 
Min.      .    .  .. 

292.5 
97.5 

322.5 
107.5 

345.0 
115.0 

360.0 
120.0 

367.5 
122.5 

WEB  STRESSES. 

Max.  stress  in  A-4'=  58.4,  in  1-6  =  53.0,  in  3-8  =  47.7, 
in  5-10=42.5,  in  7-12=37.8,  in  9-14=33.2,  in  11-16=28.7, 
in  13-18  =  24.0,  in  15-20  =  20.2,  in  17-22  =  16.3,  in  19-20r 
=  12.3,  in  21-18'  =  8.5,  in  19'-16'  =  5.4  tons. 

Min.  stress  in  A-A  =  19.5,  in  1-6  =  17.7,  in  3-8  =  15.9, 
in  5-10  =  14.2,  in  7-12  =  11.8,  in  9-14  =  9.2,  in  11-16  =  6.8, 
in  13-18  =  4.3,  in  15-20  =  1.3,  in  17-22  =  -  2.1,  in  19-20' 
=  -  5.3,  in  21-18'  =  -  8.5,  in  19'-16'  =  -  12.4  tons. 

Max.  stress  in  1-A  =  37.5,  in  A-2  =  112.5  tons. 

Min.  stress  in  l-A  =  12.5,  in  A-2  =  37.5  tons. 

We  see  from  these  results  that  the  10  diagonals  14-17 
and  17-22, 16-19  and  19-20',  18-21  and  21-18',  20-19'  and 
19'-16',  22^17'  and  17 '-14'  require  to  be  counterbraced. 

Prob.  79.  A  deck  lattice  truss,  containing  four  web  sys- 
tems, like  Fig.  39,  has  20  panels,  each  12  feet  long,  the 
depth  of  truss  is  24  feet;  the  dead  and  live  loads  are 
500  Ibs.,  and  1000  Ibs.  per  foot  per  truss :  find  the  stresses 
in  all  the  members. 


98 


BOOFS  AND  BRIDGES. 


Ans.  Max.  stresses  in  upper  chord  =  18.0,  58.5,  94.5, 126.0, 
153.0,  175.5,  193.5,  207.0,  216.0,  220.5  tons. 

Max.  stresses  in  lower  chord  =  22.5,  63.0,  99.0,  130.5, 
157.5,  180.0,  198.0,  211.5,  220.5,  225.0  tons. 

Max.  stresses  in  struts  A-3,  2-5,  etc.  =  35.0,  31.8,  28.6, 
25.5,  22.7,  19.9,  17.2,  14.4,  12.1,  9.8,  7.4,  5.1,  3.2  tons. 

Min.  stresses  in  struts  A-3,  2-5,  etc.  =  —  11.7,  — 10.6, 
-9.5,  -8.5,  -7.0,  -5.5,  -4.1,  -2.6,  -0.8,  +1.2,  +3.2, 
+  5.1,  +  7.4  tons. 

Max.  in  A-\  =  22.5  ;  min.  in  A-l  =  7.5.  Max.  in  yl-2 
=  67.5 ;  min.  in  A-2  =  22.5  tons. 

Prob.  80.  A  through  lattice  truss,  containing  four  web 
systems,  like  Fig.  39,  has  16  panels,  each  10  feet  long,  the 
depth  of  truss  is  20  feet;  the  dead  and  live  loads  are 
1000  Ibs.,  and  2000  Ibs.  per  foot  per  truss :  find  the  stresses 
in  all  the  members. 

Art.  31.  The  Post  Truss  *  —  This  truss,  shown  in 
Fig.  40,  is  a  special  form  of  the  double  triangular  truss 
(Art.  28).  The  members  3-4,  5-6,  etc.,  are  struts,  and  all 


8        10        10' 
J-'ig.  40 


the  other  diagonals  are  ties  ;  the  counters  are  shown  in 
dotted  lines.  The  distinctive  feature  of  this  truss  is  that 
the  web  struts,  or  posts,  instead  of  standing  vertically,  are 

\      *  This  truss  has  become  obsolete. 


BRIDGE  TRUSSES.  99 

inclined  by  one  half  of  a  panel  length,  and  the  ties  by  one 
and  one  half  panel  lengths.  All  the  panels  of  both  chords 
are  of  equal  length,  except  the  two  end  panels  of  the  lower 
chord,  which  are  each  one  half  a  panel  in  length. 

There  are  ambiguities  in  the  character  of  the  stresses  in 
this  truss,  as  there  are  in  all  double  systems  of  bracing. 
It  is  impossible  to  separate  the  two  systems  as  they  are 
connected  at  the  center.  But  if  we  assume  that,  under  a 
full  load,  the  members  7-10'  and  7-10  are  not  acting,  if  it 
is  a  deck  truss,  or  that  the  members  7-10',  7'-10,  9-10,  and 
9-10'  are  not  acting,  if  it  is  a  through  truss,  then  the  chord 
stresses  are  readily  found.  Also,  if  we  assume  that  the 
systems  2,  1,  6,  5,  10,  7',  8',  3',  4',  1',  2',  and  2,  1,  4,  3,  8,  7, 
10',  5',  6',  1',  2'  are  independent,  the  web  stresses  may  be 
readily  found. 

Prob.  81.  A  deck  Post  truss,  Fig.  40,  has  8  panels  in  the 
upper  chord,  each  10  feet  long,  and  20  feet  deep ;  the  dead 
and  live  loads  are  400  Ibs.,  and  1600  Ibs.  per  foot  per  truss : 
find  the  stresses  in  all  the  members. 

Dead  panel  load  per  truss  =  2  tons. 

Live  panel  load  per  truss  =  8  tons. 

tan  0  =  0.25,  and  sec  6  =  1.0308  for  the  posts. 

tan  B  =  0.75,  and  sec  0  =  1.25  for  the  ties. 

Max.  stress  in  1-3=  [2  x  £  +  1£  x  f  ]  10  =  16.3  tons. 

Max.  stress  in  4-6=  [2  x  £  +  2  x  £]  10  =  10.0  tons. 

Max.  stress  in  5-6 

=  [l£x2+(2+4+6)]  x  1.03=15.5  tons. 
Max.  stress  in  3-8 

=  [2  +  (1  +  3  +  5)]  x  1.25  =  13.8  tons. 
Min.  stress  in  3-8  =  [2  -  8  x  £]  x  1.25  =  1.3  tons. 


100  EOOFS  AND  BRIDGES. 

Thus  the  following  stresses  are  determined : 
CHORD  STRESSES. 


MEMBEBS. 

1-3 

3-5 

5-7 

7-9 

4-6 

6-3 

8-10 

10-10' 

Max.  .    . 

16.3 

28.7 

36.2 

38.7 

10.0 

25.0 

35.0 

40.0 

Min.  .    . 

3.3 

5.7 

7.2 

7.7 

2.0 

5.0 

7.0 

8.0 

STRESSES  IN  TIES. 


MEMBERS. 

1-4 

1-6 

3-8 

5-10 

7-10' 

8-9 

6-7 

4-5 

Max.  .    . 

20.6 

18.7 

13.8 

8.7 

7.5 

5.0 

2.5 

1.3 

Min.  .     . 

4.2 

3.8 

1.3 

0.0 

0.0 

0.0 

0.0 

0.0 

STRESSES  IN  POSTS. 


MEMBERS. 

1-2 

8-4 

5-6 

7-8 

9-10 

2-4 

Max.   .    . 

40.0 

20.6 

15.4 

11.4 

7.2 

0.0 

Min.    .     . 

8.0 

4.2 

3.1 

2.1 

1.0 

0.0 

Prob.  82.  A  through  Post  truss,  like  Fig.  40,  has  12 
panels  in  the  lower  chord,  each  10  feet  long  and  20  feet 
deep ;  the  dead  and  live  loads  are  1000  Ibs.,  and  2000  Ibs. 
per  foot  per  truss :  find  the  stresses  in  all  the  members. 

Art.  32.  The  B  oilman  Truss.  —  This  truss,  shown  in 
Fig.  41,  was  the  earliest  type  of  iron  truss  built  in  the 
United  States.  It  consists  of  a  series  of  inverted  king  post 
trusses  with  unequally  inclined  ties,  which  carry  the  load 
at  each  joint  directly  to  the  ends  of  the  upper  chord.  The 
vertical  pieces  are  struts;  the  upper  chord  is  in  compres- 
sion :  and  there  is  no  stress  in  the  lower  chord.  The  short 


BRIDGE  TRUSSES. 


101 


ties  shown  in  dotted  lines  in  each  panel  serve  to  stiffen  the 
structure. 

.5  7  5'      *  3' 


A  6  8  6'  A' 

Fig.  41 

Bollman  trusses  were  largely  used  in  the  Baltimore  and 
Ohio  Railroad  from  1840  to  1850 ;  but  they  were  abandoned 
long  ago. 

Prob.  83.  A  deck  Bollman  truss,  Fig.  41,  has  6  panels, 
each  12  feet  long,  and  16  feet  deep ;  the  dead  and  live  loads 
are  1000  Ibs.,  and  2000  Ibs.  per  foot  per  truss:  find  the 
stresses  in  all  the  members. 

Art.  33.  The  Fink  Truss.  —  This  truss,  shown  in 
Fig.  42,  was  first  built  about  1851,  and  was  regarded  as  an 
improvement  on  the  Bollman  truss.  It  consists  of  a  com- 

A        B       C       D        E        D'      C'       B'       A' 


bination  of  inverted  king  post  trusses  with  equally  inclined 
ties,  as  the  primary  system  AdA',  the  secondary  systems 
AbE  and  A'b'E,  and  the  tertiary  systems  AaC,  CcE,  etc. 
The  vertical  pieces  are  struts,  the  upper  chord  is  in  com- 
pression, and  there  is  no  stress  in  the  lower  chord ;  all  the 
diagonals  are  ties. 

The  load  may  be  carried  either  upon  the  upper  or  the 
lower  chord,  though  it  has  been  used  more  often  for  deck 


102  ROOFS  AND  BRIDGES. 

than  for  through  bridges.  These  trusses  were  built  from 
1851  down  to  about  1876 ;  but  they  are  not  now  generally 
regarded  with  favo*  by  bridge  builders. 

Prob.  84.  A  deck  Fink  truss,  Fig.  42,  has  8  panels,  each 
12  feet  long,  and  16  feet  deep ;  the  dead  and  live  loads  are 
500  Ibs.,  and  2000  Ibs.  per  foot  per  truss :  find  the  stresses 
in  all  the  members. 

The  dead  panel  load  per  truss  =  3  tons. 

The  live  panel  load  per  truss  =  12  tons. 

We  see  at  once,  from  Fig.  42,  that  the  maximum  stresses 
in  all  the  members  occur  when  there  is  a  full  panel  load  at 
every  apex.  Hence  the  maximum  stresses  will  be  when  15 
tons  are  placed  at  every  apex,  and  the  minimum  stresses 
will  be  one  fifth  of  the  maximum.  The  maximum  stress  in 
each  of  the  posts  Ba,  DC  is  — 15  tons.  The  maximum 
stress  in  the  post  Cb  is  the  panel  load  of  15  tons  at  C,  plus 
one  half  the  panel  load  of  15  tons  at  D,  plus  one  half  the 
panel  load  of  15  tons  at  B,  =  —  30  tons.  Similarly  the 
stress  in  Ed  =  —  60  tons. 

Also, 

Stress  in  Aa  =  \  x  15  sec  0  =  9.4  tons  =  Ca  =  Cc  =  EC. 
Stress  in  Ab  =  $  x  30  sec  el  =  27.0  tons  =  Eb. 
Stress  in  Ad  =  $  x  60  sec  02  =  95.0  tons. 
Stress  in  AA1  =  -  [|  x  15  x  f  +  $  x  30  x  f  +  £  X  60  x  J£| 
=  —118.1  tons. 

Prob.  85.  A  deck  Fink  truss  has  8  panels,  each  10  feet 
long  and  15  feet  deep ;  the  dead  and  live  loads  are  400  Ibs. 
and  2000  Ibs.  per  foot  per  truss :  find  the  stresses  in  all  the 
members.  - 


BRIDGE  TRUSSES.  108 


BRIDGE  TRUSSES  WITH  INCLINED  CHORDS. 

Art.  34.  The  Parabolic  Bowstring  Truss.  —  In  this 
truss,  Fig.  43,  the  lower  chord  is  horizontal,  and  the  upper 


A  E       D       C 

Xlw  Fi6. 43  *lw 

chord  joints  lie  on  the  arc  of  a  parabola;  the  verticals  may 
be  in  compression  and  the  diagonals  in  tension,  or  the  ver- 
ticals may  be  in  tension  and  the  diagonals  in  compression. 

Let  I  =  the  length  of  the  span  AB,  d  =  the  height  of  the 
arc  OC  at  the  center  of  the  span,  and  w  =  the  uniform  load 
in  Ibs.  per  foot  of  truss ;  let  (x,  y)  be  any  joint  P  of  the 
upper  chord  referred  to  the  axes  AB  and  AY,  and  8  the 
stress  in  the  lower  chord  panel  opposite  P.  Then  we  have 


(1) 


The  equation  of  the  parabola  referred  to  its  vertex  0  as 
origin  is 

(il-xY  =  2p(d-y) (2) 

72  72 

and  for  the  point  A  this  becomes  —  =  2pd;     .-.  2p=—. 

4  4d 

Substituting  this  in  (2)  and  solving  for  y,  we  get 

y  =  *j(lx-x>)      ......     (3) 

which  in  (1)  gives  S  =  —  .    .  ' (4) 

Hence,  for  a  uniform  load  on  a  parabolic  truss  the  stress  in 
the  lower  chord  is  the  same  in  every  paneL 


104 


HOOFS   AND   BRIDGES. 


Since  the  horizontal  component  of  the  stress  in  any 
diagonal,  as  PD  for  example,  is  equal  to  the  difference  of 
the  chord  stresses  in  the  adjacent  panels,  ED  and  DC. 

Therefore,  for  a  uniform  load  there  is  no  stress  in  any 
diagonal. 

It  follows  at  once  that,  for  a  uniform  load  the  horizontal 
component  of  the  stress  in  the  upper  chord  is  the  same  in  every 
panel,  and  is  expressed  by  equation  (4). 

Therefore,  for  a  uniform  load  the  stress  in  any  inclined 
panel  of  the  upper  chord  is  equal  to  the  stress  in  a  panel  of  the 
lower  chord  multiplied  by  the  secant  of  the  inclination. 

Prob.  86.  A  parabolic  bowstring,  Fig.  44,  as  a  through 
bridge,  has  8  panels,  each  10  feet  long,  and  10  feet  center 


depth,  the  verticals  take  either  tension  or  compression,  the 
diagonals  are  ties ;  the  dead  and  live  loads  are  400  Ibs.,  and 
800  Ibs.  per  foot  per  truss :    find  the  stresses  in  all  the 
members. 
The  max.  stress  in  any  panel  of  the  lower  chord,  by  (4), 


1200  x  80' 


80  x  2000 
Min.  stress  =     x  48 


48  tons. 

16  tons. 


Dead  panel  load  =  2  tons. 
Live  panel  load  =  4  tons. 


BRIDGE  TRUSSES.  105 

By  (3)  we  have  lengths  of  3-4,  5-6,  and  7-8  =  4.375,  7.5, 
and  9.375  feet. 


Max.  stress  in  2-3  =  48  *  V  <10^  +  <4-375>'  =  52.4  tons. 

Since  the  dead  load  produces  no  stresses  in  the  diagonals, 
the  max.  stress  in  any  diagonal  is  found  by  putting  only  the 
live  load  on  the  bridge  in  the  position  to  give  the  max. 
shear  in  that  member  (Art.  22). 

Thus,  for  the  maximum  stress  in  5-8  the  live  load  is 
placed  on  the  right,  the  center  of  moments  is  on  the  lower 
chord  at  20  feet  to  the  left  of  the  point  2.  The  lever  arm 
of  5-8  is  30  feet. 

The  reaction  for  this  loading 

=  |(1  +  2  +  3  +  4  +  5)  =  7.5  tons. 

.-.  Stress  in  5-8  =  7'5  *  20  =  5  tons. 

oU 

For  live  load  on  the  left,  the  stress  in  5-8  =  0,  and  the 
counter  6-7  comes  into  action.     The  lever  arm  of  6-7  is 
27.4  feet  ;  the  reaction  for  this  loading  =  6.5  tons.     There- 
fore the  stress  S  for  6-7  is  found  from  the  equation 
6.5  x  20  -  4  (30  +  40)+  S  x  27.4  =  0. 
.-.  S  =  stress  in  6-7  =  5.5  tons. 

If  we  cut  a  vertical  as  5-6  by  a  section,  place  the  live 
load  on  the  right,  and  take  the  center  of  moments  at  the 
intersection  of  3-5  and  4-6,  which  is  4  feet  to  the  left  of 
the  point  2,  we  shall  obtain  the  maximum  compression  in  it 
caused  by  the  live  load.  Thus, 

The  reaction  for  this  loading  =  7.5  tons. 

.-.  Max.  stress  in  5-6  due  to  live  load 


=  -1.25  tons. 


Min.  stress  in  5-6  =  2.0  -  1.25  =  0.75  tons. 


106 


ROOFS  AND  BRIDGES. 


The  maximum  stress  in  each  vertical  is  one  of  tension, 
and  will  occur  when  the  live  load  covers  the  bridge,  and  is  a 
dead  and  live  panel  load,  or  6  tons. 

If  there  be  110  live  load  on  the  truss,  the  stress  in  any 
vertical  is  a  dead  panel  load,  or  a  tension  of  2  tons.  Now 
let  the  live  load  come  on  from  the  right,  a  part  of  it  which 
goes  to  the  left  support  will  cause  compression  in  each 
vertical,  or  will  diminish  the  tension  in  that  vertical  due 
to  the  dead  load.  Thus,  when  the  live  load  reaches  the 
joint  8',  it  will  cause  in  the  vertical  9-10  a  compression  of 
2.2  tons,  which  will  neutralize  the  dead  load  tension  of 
2  tons,  and  leave  as  the  result  a  compression  of  0.2  tons. 
Also,  when  it  reaches  the  apex  8,  it  will  cause  in  the  vertical 
5-6  a  compression  of  1.25  tons,  as  we  saw  above,  which  will 
neutralize  that  much  of  the  dead  load  tension  of  2  tons,  and 
leave  as  the  result  a  tension  of  0.75  tons. 

The  following  stresses  are  found  in  a  manner  similar  to 
the  above : 

CHORD  STRESSES. 


MEMBERS. 

2-3 

3-5 

5-7 

7-9 

2-4,  4-6,  etc. 

Max.  .  .  . 
Min.    .  .  . 

-52.4 
-17.5 

-50.3 
-16.8 

-48.8 
-16.3 

-48.1 
-16.0 

+  48.0 
+  16.0 

STRESSES  IN  DIAGONALS. 


MEMBERS. 

8-6 

5-8 

7-10 

4-5 

6-7 

8-9 

Max.   .  .  . 

+4.3 

+  5.0 

+  5.5 

+  5.0 

+6.5 

+  6.7 

Min.     .  .  . 

0.0 

0.0 

0.0 

0.0 

0.0 

0.0 

BRIDGE  TRUSSES. 


107 


STRESSES  IN  VERTICALS. 


MEMBERS. 

3-t 

5-6 

1-8 

9-10 

Max                      .      ... 

+6.0 

+60 

+  6  0 

+6  0 

Min  

2.0 

+  0.8 

+0.0 

-0.2 

Prob.  87.  A  parabolic  bowstring  as  a  through  bridge, 
Fig.  44,  has  8  panels,  each  10  feet  long  and  10  feet  center 
depth ;  the  verticals  are  ties  and  the  diagonals  are  braces ; 
the  dead  and  live  loads  are  400  Ibs.  and  1600  Ibs.  per  foot 
per  truss :  find  the  stresses  in  all  the  members. 

Ans.   Max.  stress  in  each  panel  of  lower  chord  =  +80  tons. 
Max.  stresses  in  upper  chord 

=  _  87.3,  -  83.8,  -  81.4,  -  80.2  tons. 
Max.  stress  in  3-6 

=  -  8.7,  in  5-8  =  -  10.0,  in  7-10  =  -  10.9,  in  4-5 
=  -  10.0,  in  6-7  =  -  10.9,  in  8-9  =  -  11.3  tons. 
Max.  stresses  in  verticals  . 

=  +10.0,  +12.5,  +-14.0,  +  14.5  tons. 
Min.  stresses  in  verticals  =+-2.0,  +2.0,  +2.0,  +2.0  tons. 
Min.  stresses  in  diagonals  =  0. 

Prob.  88.  A  through  parabolic  bowstring  truss,  Fig.  44, 
has  8  panels,  each  12  feet  long  and  16  feet  center  depth ;  the 
verticals  are  ties  and  the  diagonals  are  braces ;  the  dead  and 
live  loads  are  500  Ibs.  and  2000  Ibs.  per  foot  per  truss :  find 
the  stresses  in  all  the  members. 

Ans.   Max.  stress  in  each  panel  of  lower  chord  =  +  90  tons. 

Min.  stress  in  each  panel  of  lower  chord  =  +  18  tons. 

Max.  stresses  in  upper  chord 

=  -  104.1,  -  97.5,  -  92.7,  -  90.3  tons. 


108  ROOFS  AND  BRIDGES. 

Min.  stresses  in  upper  chord 

=  _20.8,  -19.5,  -18.5,  -18.1  tons. 
Max.  stresses  in  verticals 

=  +  15.0,  +  18.7,  +  21.0,  +  21.7  tons. 
Min.  stresses  in  verticals  =  +3.0,  +3.0,  +3.0,  +3.0  tons. 
Max.  stress  in  3-6 

=  -  10.4,  in  5-8  =  - 12.7,  in  7-10  =  -  14.4,  in  4-5 

=  -  12.7,  in  6-7  =  -  14.4,  in  8-9  =  -  15.0  tons. 
Min.  stresses  in  diagonals  =  0. 

Prob.  89.  A  through  parabolic  bowstring  truss  has  12 
panels,  each  8  feet  long,  and  12  feet  center  depth;  the 
verticals  are  ties  and  the  diagonals  are  braces;  the  dead 
and  live  loads  are  500  Ibs.  and  2000  Ibs.  per  foot  per  truss : 
find  the  stresses  in  all  the  members. 

Art.  35.  The  Circular  Bowstring  Truss.  —  This  form 
of  truss,  Fig.  45,  is  often  used  for  highway  bridges.  The 


joints  of  the  upper  chord  lie  upon  the  arc  of  a  circle,  each 
joint  being  directly  over  the  middle  of  the  panel  below. 
The  diagonals  are  built  to  take  either  tension  or  com- 
pression. 

Prob.  90.  A  through  circular  bowstring  truss,  Pig.  45,  has 
panels,  each  12  feet  long,  and  11.7  feet  center  depth ;  the 
joints  of  the  upper  chord  lie  on  the  arc  of  a  circle  of  60  feet 
radius ;  the  dead  and  live  loads  are  450  Ibs.  and  1360  Ibs. 
per  foot  per  truss :  find  the  stresses  in  all  the  members. 


BRIDGE   TRUSSES.  109 

Ans.  Max.  stress  in  2-3  =  —  49.4,  in  3-5  =  —  54.3,  in 
5-7  =  -  51.1,  in  7-7'  =  -  50.1  tons. 

Max.  stress  in  2-4  =  +  41.1,  in  4-6  =  +  45.8,  in  6-8 
=  +  47.3  tons. 

Max.  stress  in  3-4  =  +  9.9,  in  4-5  =+  9.6,  in  5-6 
=  +  10.4,  in  6-7  =  +  10.0,  in  7-8  =  +  10.8  tons. 

Min.  stress  in  4-5  =  —  1.6,  in  5-6  =  —  1.3,  in  6-7 
=  —  3.6,  in  7-8  =  —  3.1  tons. 

This  problem  is  taken,  with  slight  changes,  from  Burr's  Stresses 
in  Bridge  and  Roof  Trusses. 

Prob.  91.  A  through  circular  bowstring  truss  has  8 
panels,  each  15  feet  long,  the  center  depth  is  19.74  feet; 
the  joints  of  the  upper  chord  lie  on  the  arc  of  a  circle  of 
100  feet  radius,  each  joint  being  directly  over  the  center 
of  the  panel  below ;  the  dead  and  live  loads  are  1000  Ibs., 
and  2000  Ibs.  per  foot  per  truss :  find  the  stresses  in  all  the 
members. 

Art.  36.  Snow  Load  Stresses.  —  For  railway  bridges 
the  snow  load  is  not  taken  into  account,  since  the  floor  is 
open,  so  that  but  little  is  retained.  For  highway  bridges 
the  snow  load  is  taken  from  0  to  15  Ibs.  per  square  foot  of 
floor  surface,  depending  upon  the  climate  where  the  bridge 
is  situated.*  The  snow  load  is  taken  lower  than  for  roofs, 
since  the  full  live  load  is  not  likely  to  come  upon  the  bridge 
while  it  is  heavily  loaded  with  snow.  Since  the  snow  load 
is  uniform,  the  stresses  due  to  it  are  computed  in  the  same 
way  as  the  dead  load  stresses ;  or  the  snow  load  and  dead 
load  stresses  are  proportional  to  the  corresponding  apex 
loads  (Art.  8). 

*  In  building  highway  bridges  in  England  and  France  the  snow  load  is 
not  generally  considered. 


110  ROOFS  AND  BRIDGES. 

Prob.  92.  A  through  Howe  truss,  Fig.  32,  has  10  panels, 
each  12  feet  long  and  12  feet  deep ;  the  width  of  roadway 
is  20  feet,  and  the  width  of  each  sidewalk  is  5  feet;  the 
snow  load  is  10  Ibs.  per  square  foot  of  floor :  find  the  snow 
load  stresses  in  all  the  members. 

Ans.    Stresses  in  lower  chord 
=  4.1,  7.2,  9.5,  10.8,  11.3  tons. 

Stresses  in  the  verticals  =  4.1,  3.2,  2.3,  1.4,  0.9  tons. 

Stresses  in  the  diagonals 

=  -5.7,  -4.4,  -3.2,  -1.9,  -0.6  tons. 

Prob.  93.  A  deck  Pratt  truss,  Fig.  34,  has  10  panels,  each 
16  feet  long  and  16  feet  deep ;  the  width  of  roadway,  in- 
cluding sidewalks,  is  35  feet;  the  snow  load  is  15  Ibs.  per 
square  foot  of  floor :  find  the  snow  load  stresses  in  all  the 
members. 

Art.  37.  Stresses  due  to  Wind  Pressure.  —  In  bridges 
of  large  span,  it  is  often  found  that  the  stresses  produced  in 
some  of  the  members  by  a  gale  of  wind  are  almost  as  great 
as  i^iose  caused  by  both  the  dead  and  live  loads.  In  the 
principal  members  of  the  Forth  bridge,  Scotland,  Sir  Ben- 
jamin Baker  estimated  that  the  maximum  stresses  due  to 
these  three  separate  forces  were  as  follows : 

Stress  due  to  dead  load  =  2282  tons. 

Stress  due  to  live  load    =  1022  tons. 

Stress  due  to  wind  load  =  2920  tons. 

In  estimating  the  wind  pressure,  and  the  resulting  stresses 
in  the  members  of  a  bridge,  the  practice  of  engineers  varies 
greatly.  Different  railroads  have  their  own  specifications 
for  wind  pressure.  The  standard  wind  pressure  per  square 
Joot  ranges  in  this  country  from  30  to  50  Ibs.  It  is  assumed 
by  many  engineers  that  there  will  be  no  train  on  the  bridge 


BRIDGE   TRUSSES.  Ill 

when  the  wind  blows  with  greater  pressure  than  30  Ibs.  per 
square  foot.  A  wind  pressure  of  30  Ibs.  per  square  foot  will 
overturn  an  empty  freight  car.  In  such  a  gale  it  would  be 
hardly  possible  for  a  train  to  reach  the  bridge.  It  might, 
however,  be  caught  there  by  a  sudden  squall ;  and  this  is 
just  what  appears  to  have  happened  in  the  case  of  the  Tay 
bridge,  which  was  destroyed  by  a  wind  storm  in  1879.  A 
maximum  pressure  of  50  Ibs.  per  square  foot  is  taken  as 
applying  to  the  bridge  alone. 

The  surface  exposed  to  the  wind  action  is  commonly 
taken  as  double  the  side  elevation  of  one  truss,  for  the 
reason  that  the  windward  truss  cannot  afford  much,  shelter 
to  the  leeward  truss,  whatever  may  be  the  direction  of  the 
wind.  In  a  heavy  gale  of  wind  there  is  not  much  shelter  to 
be  found  under  the  lee  of  a  lamp-post  at  a  distance  of 
20  feet  from  it,  even  if  the  post  be  directly  to  windward. 
Experiments  show  that  the  wind  pressure  against  the  two 
trusses  of  a  bridge  is  more  than  1.8  times  that  on  the 
exposed  surface  of  one  truss.. 

For  Highway  Bridges  the  wind  pressure  is  frequently 
taken  at  about  30  Ibs.  per  square  foot  of  exposed  surface  of 
both  trusses.  It  is  assumed  that  there  will  be  no  live  load 
upon  the  bridge  when  the  wind  is  blowing  at  this  maximum 
pressure  of  30  Ibs.  per  square  foot. 

For  Railroad  Bridges  the  wind  pressure  is  taken  at 
about  30  Ibs.  per  square  foot  of  exposed  surface  of  both 
trusses,  and  about  300  Ibs.  per  linear  foot  due  to  the  train 
surface.  The  train  surface  is  about  10  square  feet  for  each 
linear  foot  of  the  bridge.  The  30  Ibs.  per  square  foot  of  the 
trusses  is  treated  usually  as  a  dead  load,  and  the  300  Ibs. 
per  linear  foot  due  to  the  train  surface  as  a  live  load.  This 
regards  each  truss  as  fully  exposed  even  when  the  train  is 
on,  though  it  partially  shelters  one  truss. 


112  ROOFS  AND  BRIDGES. 

To  estimate  the  30  Ibs.  pressure  per  square  foot  of  exposed 
surface  of  both  trusses,  when  this  surface  is  not  known, 
it  is  customary  now  to  use  the  following  rule :  Take  150 
Ibs.  per  linear  foot  per  truss,  or  75  Ibs.  per  linear  foot  for  each 
chord. 

The  wind  load  on  the  trusses  is  assumed  to  be  divided 
equally  between  the  upper  and  lower  lateral  trusses,  while 
the  wind  load  on  the  train  is  all  taken  by  the  lateral  truss 
belonging  to  the  loaded  chord.  The  lateral  trusses  are  the 
Lorizontal  trusses  placed  between  the  chords  of  the  vertical 
trusses.  The  chords  of  the  vertical  trusses  are  also  the 
chords  of  the  lateral  trusses.  The  lateral  trusses  of  a  bridge 
are  either  of  the  Pratt,  Howe,  or  Warren  type,  correspond- 
ing generally  with  the  type  of  the  main  trusses. 

REM. —  In  finding  wind  stresses  the  loads  on  the  lateral  system 
of  the  unloaded  chord  are  considered  as  applied  equally  upon  the  two 
sides,  windward  and  leeward,  while  the  loads  on  the  lateral  system  of 
the  loaded  chord  are  considered  as  applied  wholly  on  the  windward 
side.  The  stresses  are  then  readUy  found  by  methods  already  familiar 
for  finding  dead  load  and  live  load  stresses. 

The  lower  lateral  system  in  a  through  bridge  and  the  upper  lateral 
system  in  a  deck  bridge  are  calculated  for  a  dead  load  of  30  Ibs.  per 
square  foot  of  exposed  surface  of  both  trusses,  and  a  live  load  of 
300  Ibs.  per  linear  foot  of  train ;  while  the  other  lateral  system  is 
calculated  only  for  a  dead  load  of  30  Ibs.  per  square  foot  of  exposed 
surface  of  both  trusses. 

The  resulting  chord  stresses  should  be  combined  with 
those  due  to  the  dead  load,  or  with  those  due  to  the  dead 
and  live  loads  if  the  live  load  acts  with  the  wind  load.  The 
lower  chord  is  always  in  tension  under  the  action  of  the 
dead  and  live  loads.  When  the  wind  acts  the  bridge  is  bent 
laterally,  and  the  windward  lower  chord  has  its  maximum 
tension  diminished  by  the  compression  due  to  the  wind, 


BRIDGE  TRUSSES. 


113 


while  the  lower  leeward  chord  has  its  tension  increased  by 
the  tension  due  to  the  wind.  The  compression  in  the  wind- 
ward chord  due  to  the  wind  may  exceed  the  tension  due  to 
the  dead  load,  or  to  the  dead  and  live  loads  combined.  The 
lower  chord  on  each  side  then  should  not  only  be  able  to 
sustain  the  total  maximum  of  dead,  live,  and  wind  loads,  but 
to  act  as  a  strut  to  resist  compression. 

Prob.  94.  A  through  Pratt  truss  railway  bridge  16  feet 
wide,  half  of  which  is  shown  in  Fig.  47,  has  12  panels,  each 
16  feet  long ;  the  upper  and  lower  lateral  systems,  Figs.  46 


If       lh 


IB      ic      in     JE      IF      JH 


IXIXIX1X1XM; 

A'  B'         C'         D'         E'         F*        H' 


D'         E' 
Fig.  48 


and  48,  are  Pratt  trusses :  find  the  stresses  in  both  lateral 
systems  due  to  a  wind  pressure  of  30-lbs.  per  square  foot  of 
j  _.._r__  _*  botll  trusses  anci  300  Ibs.  per  linear  foot 


exposed  surface  of 
of  train  surface. 


114  ROOFS  AND  BRIDGES. 


(1)  UPPER  LATERAL  SYSTEM  (10  PANELS). 

Panel  load  for  one  chord  (see  rule)  =  ^^  =  0.6  tons. 

.2000 

This  load  acts  as  a  dead  load  at  each  apex  of  the  wind- 
ward and  leeward  chords  in  the  direction  of  the  arrows, 
shown  in  Fig.  46;  hence  there  is  no  stress  in  the  dotted 
diagonals. 

tan0  =  l;    sec  0  =  1.414. 

Stress  in  hh'  =  —  0.6  tons. 
Stress  in  ff  =  -  0.6  x  2  =  -  1.2  tons. 
Stress  in  ee'  =  -  0.6  x  4  =  -  2.4  tons. 
Stress  in  be  =  -  0.6  x  9  x  1  =  -  5.4  tons. 
Stress  in  cd  =  -[5.4  -f  4.2]  x  1  =  -  9.6  tons. 
Stress  in  fh'  =  +  0.6  x  1.414  =  +  0.8  tons. 
Stress  in  ef  =  +  3  x  0.6  x  1.414  =  +  2.5  tons. 

(2)  LOWER  LATERAL  SYSTEM  (12  PANELS). 
Panel  load  for  both  chords 


The  former  we  treat  as  a  dead  load,  the  latter  as  a  live 
load,  both  acting  at  each  apex  of  the  windward  chord  in  the 
direction  of  the  arrows,  shown  in  Fig.  48.  (See  Kemark.) 

Stress  in  AB  =  —  5.5  x  3.6  =  —  19.8  tons. 
Stress  in  CO 

=  -j~4.5xl.2+||(l+2+3  +  •••  +  10)1  =-16.4  tons. 

Stress  in  DE' 

=  [2.5x1.2+0.2(1+2+3+  -  +8)]  1.414=  14.4  tons. 


BRIDGE   TRUSSES.  115 

Thus  the  following  stresses  are  computed : 

UPPER  LATERAL  SYSTEM. 

Stresses  in  chords 

=  _  5.4,  -  9.6,  -  12.6,  -  14.4,  -  15.0  tons. 

Stresses  in  struts 

=  -  5.7,  -  4.8,  -  3.6,  -  2.4,  -  1.2,  -  0.6  tons. 

Stresses  in  diagonals 

=  +  7.6,  +  5.9,  +  4.2,  +  2.5,  +  0.8  tons. 

LOWER  LATERAL  SYSTEM. 

Stresses  in  chords  =  19.8,  36.0,  48.6,  57.6,  63.0,  64.8  tons. 
Stresses  in  struts 

=  -21.6,  -19.8,  -16.4,  -13.2,  -10.2,  -7.4,  -5.4 tons. 
Stresses  in  diagonals  =  27.9,  23.1, 18.6, 14.4, 10.4,  6.8  tons. 

The  chord  bcdh  is  in  compression  under  the  action  of  the 
dead  and  live  loads ;  this  compression  is  increased  by  that 
due  to  the  wind  pressure,  as  just  found. 

The  chord  A'B'C'H'  is  in  tension  under  the  action  of  the 
dead  and  live  loads ;  this  tension  is  increased  by  that  due  to 
the  wind  pressure,  as  just  found. 

When  the  wind  blows  on  the  opposite  side  of  the  bridge 
the  diagonal  and  chord  stresses  are  to  be  interchanged, 
while  the  strut  tresses  remain  the  same.  Thus  there  will 
be  the  same  stresses  in  &'c',  c'd',  etc.,  as  in  be,  cd,  etc.,  and 
the  same  in  A'B',  B'C',  etc.,  as  in  AB,  BC,  etc.,  and  the 
dotted  system  of  braces  will  act  in  each  lateral. 

The  wind  loads  on  the  left  half  of  the  upper  lateral  are 
transferred  to  the  abutment  at  A  by  means  of  the  portal 
bracing  AA'b'b  in  the  transverse  plane  of  Ab. 


116  ROOFS  AND  BRIDGES. 

Prob.  95.  A  through  Howe  truss  railway  bridge  20  feet 
wide  has  12  panels,  each  20  feet  long;  the  upper  and  lower 
lateral  systems  are*  Howe  trusses  :  find  the  stresses  in  both 
lateral  systems  due  to  the  same  wind  pressure  as  in  the 
previous  problem. 

By  taking  the  dotted  diagonals,  Figs.  47,  46,  and  48  will 
represent  the  left  half  of  the  Howe  truss  and  the  upper  and 
lower  lateral  systems.  Then  by  rule  : 

Panel  load  for  one  chord  of  upper  lateral  =  0.75  tons,  to 
be  taken  as  a  dead  load. 

Panel  load  for  both  chords  of  lower  lateral 

=  1.5  tons  +  3.0  tons ; 

the  former  to  be  taken  as  a  dead  load,  the  latter  as  a  live 
load,  all  acting  in  the  direction  of  the  arrows. 

Stress  in  b 'c  —  -  9  x  .75  x  1.414  =  -  9.5  tons. 
Stress  in  C'D 

=  -  [3.5  x  1.5  +  & (1  +  2  +  ...  +  9)]  1.414  =  -  23.3  tons. 
Stress  in  B'C'  =  10  x  4.5  x  1  =  45  tons. 

UPPER  LATERAL  SYSTEM. 

Stresses  in  chords  =  6.8,  12,0,  15.8,  18.0,  18.8  tons. 
Stresses  in  struts  =  0.4,  6.0,  4.5,  3.0,  1.5,  0.8  tons. 
Stresses  in  diagonals 

=  -  9.5,  -  7.4,  -  5.3,  -  3.2,  -  1.1  tons. 

LOWER  LATERAL  SYSTEM. 

Stresses  in  chords  =  24.8,  45.0,  60.8,  72.0,  78.8,  81.0  tons. 
Stresses  in  struts  =  2.3,  20.5,  16.5,  12.8,  9.2,  6.0,  3.8  tons. 
Stresses  in  diagonals 

=  _  34.9,  -  28.9,  -  23.3,  -  18.0,  -  13.0,  -  8.5  tons. 


BRIDGE   TRUSSES.  117 

Prob.  96.  A  deck  Pratt  truss  railroad  bridge  20  feet  wide 
has  10  panels,  each  20  feet  long  ;  the  upper  and  lower  lateral 
systems  are  Pratt  trusses  :  find  the  stresses  in  tfoth  lateral 
systems  due  to  a  wind  pressure  of  40  Ibs.  per  square  foot  of 
exposed  surface  of  both  trusses  and  300  Ibs.  per  linear  foot 
of  train  surface. 

By  taking  only  the  full  diagonals,  Fig.  46  will  represent 
the  left  half  of  the  deck  Pratt  truss  and  also  the  upper  and 
lower  lateral  systems. 

Since  this  is  a  deck  bridge,  the  upper  laterals  are  on  the 
loaded,  and  the  lower  laterals  are  on  the  unloaded  chord. 
Hence  by  the  rule  : 

Panel  load  for  one  chord  on  lower  lateral  system 
100  x  20         , 


to  be  taken  as  a  dead  load  at  each  apex  of  the  windward 
and  leeward  chords  (Rem.). 
Panel  load  for  both  chords  on  upper  lateral  system 


the  former  to  be  taken  as  a  dead  load,  the  latter  as  a  live 
load,  both  acting  at  each  apex  of  the  windward  chord  (Rem.), 
and  all  acting  in  the  direction  of  the  arrows  in  Fig.  46. 

UPPER  LATERAL  SYSTEM. 

Stresses  in  chords 

=  _  22.5,  -  40.0,  -  52.5,  -  60.0,  -  62.5  tons. 
Stresses  in  struts 

=  _25.0,  -22.5,  -17.8,  -13.4,  -  9.3,  -  5.5  tons. 
Stresses  in  diagonals 

=  +  31.7,  +25.1,  +18.9,  +13.1,  +7.8  tons. 


118  ROOFS  AND  BRIDGES. 

LOWER  LATERAL  SYSTEM. 

Stresses  In  chords 

=  _  9.0,  - 16.0,  -  21.0,  -  24.0,  -  25.0  tons. 
Stresses  in  struts 

=  _  9.5,  _  8.0,  -  6.0,  -  4.0,  -  2.0,  - 1.0  tons. 
Stresses  in  diagonals 

=  + 12.7,  +  9.9,  +  7.1,  +  4.2,  + 1.4  tons. 

Prob.  97.  A  through  Pratt  truss  railroad  bridge  16^  feet 
wide  has  9  panels,  each  17  feet  long ;  the  upper  and  lower 
lateral  systems  are  Pratt  trusses :  find  the  stresses  in  both 
lateral  systems  due  to  the  same  wind  pressures  as  in 
Prob.  94. 

Art.  38.  The  Factor  of  Safety  for  a  body  is  the  ratio 
of  its  breaking  to  its  working  stress ;  or  it  is  the  ratio  of 
the  load  which  will  just  crush  the  body  to  the  assumed 
load,  or  load  which  it  is  intended  to  carry.  Thus,  if  the 
tearing  unit-stress  of  an  iron  plate  be  20  tons  and  the  work- 
ing unit-stress  be  4  tons,  the  factor  of  safety  will  be  5. 

The  value  of  the  factor  of  safety  is  generally  assumed  by 
the  engineer ;  different  engineers  assume  different  factors  of 
safety,  depending  somewhat  upon  the  manner  of  applying 
the  loads,  the  character  of  the  structure,  and  the  nature  and 
quality  of  the  material.  Thus,  for  steady  loads  and  slowly 
varying  stresses,  the  factor  of  safety  may  be  low ;  but  when 
the  load  is  applied  with  shocks  and  sudden  stresses,  the 
factor  ought  to  be  large.  In  a  building  the  stresses  on  the 
walls  are  steady,  and  hence  the  factor  of  safety  may  be  low. 
In  a  bridge  the  stresses  on  the  different  members  are  more 
or  less  varying,  and  hence  the  factor  of  safety  must  be 
higher. 


BRIDGE   TRUSSES.  119 

Also,  it  has  been  seen  (Art.  16)  that  the  live  load  for 
a  short  span  is  much  more  than  that  for  a  long  span.  For 
this  reason  the  variation  of  stress  in  passing  from  a  loaded 
to  an  unloaded  state  is  much  greater  in  the  members  of  a 
short  span  than  in  those  of  a  long  one.  Consequently,  the 
material  in  a  short  span  will  suffer  what  is  termed  fatigue 
more  than  in  a  long  one.  And  although  the  fatigue  of 
metals  is  a  subject  not  yet  well  understood,  yet  it  is  clearly 
established  that  these  sudden  changes  of  stress  in  short 
spans  demand  a  larger  factor  of  safety  than  those  in  long 
spans. 

In  American  practice  the  values  of  the  factor  of  safety, 
for  steady  and  for  varying  stresses,  are :  from  3  to  7  for 
wrought  iron,  from  5  to  15  for  cast  iron,  from  4  to  8  for 
steel,  from  8  to  12  for  timber,  and  from  12  to  30  for  stone 
or  brick. 

At  times  it  may  be  necessary  to  employ  a  much  larger 
factor  of  safety  than  either  of  these,  owing  to  local  circum- 
stances. If  the  risk  attending  failure  (such  as  loss  of  life 
or  property)  is  small,  the  factor  of  safety  may  be  small. 
But  if  the  risk  is  large,  the  factor  of  safety  must  be  large 
in  proportion.  With  a  bridge  in  perfect  condition,  a  very 
small  factor  would  be  sufficient.  But  no  bridge  is  in  per- 
fect condition;  all  rough  places,  such  as  rail  joints,  more 
or  less  open,  produce  shocks  which  cause  sudden  stresses. 
These  stresses  cannot  be  measured.  A  very  large  allowance 
has  to  be  made  for  these  uncertainties  and  for  the  imper- 
fect state  of  our  knowledge ;  and  therefore  there  must  be 
a  large  factor  of  safety  to  cover  all  uncertainties. 


CHAPTER   III. 

BRIDGE   TRUSSES   WITH  UNEQUAL  DISTRIBUTION 
OF  THE   LOADS. 

Art.  39.  Preliminary  Statement.  —  The  preceding 
chapter  has  treated  of  stresses  produced  by  dead  loads  and 
uniformly  distributed  live  loads.  While  this  is  the  general 
method  of  treatment  for  highway  bridges,  and  in  English 
practice  for  railway  bridges,  it  has  become  the  general 
practice  of  American  engineers  to  calculate  these  stresses 
for  railway  bridges  by  one  of  the  following  three  methods : 

(1)  The  use  of  a  uniformly  distributed  excess  load  covering 
one  or  more  panels  followed  by  a  uniform  train  load  cover- 
ing the  whole  span. 

(2)  The  use  of  one  or  two  concentrated  excess  loads  with 
a  uniform  train  load  covering  the  span. 

(3)  The  use  of  the  actual  specified  locomotive  wheel  loads 
followed  by  a  uniform  train  load. 

It  is  proposed  in  this  chapter  to  show  how  to  find  the 
maximum  stress  in  each  member  of  a  truss  by  each  of  these 
three  methods. 

Art.  40.  Method  of  Calculating  Stresses  when 
the  Uniform  Train  Load  is  preceded  by  one  or 
more  Heavy  Excess  Panel  Loads.  —  This  method  of 
finding  maximum  stresses  is  sometimes  used  to  avoid  the 
laborious  practice  of  finding  the  stresses  due  to  the  actual 
locomotive  wheel  loads  to  be  described  later. 
120 


BRIDGE  TRUSSES.  121 

Prob.  98.  A  through  double  Warren  truss,  Fig.  49,  has 
10  panels,  each  12  feet  long  and  12  feet  deep;  the  dead 
load  is  1000  Ibs.  per  foot  per  truss,  and  the  train  load  is 

1         3        5          7          9        11       <J         7'       5'       3'         l' 

»*~      ^        g         £        n>       12      10'      8'       6'       t'       *f* 
Trig.  40 

2000  Ibs.  per  foot  per  truss,  preceded  by  one  locomotive 
panel  load  of  30  tons  per  truss  :  find  the  stresses  in  all  the 
members. 

We  may  first  find  the  stresses  caused  by  the  uniform 
dead  and  train  loads,  and  then  the  stresses  caused  by  the 
excess  locomotive  load,  and  add  the  results  ;  or,  we  may 
determine  the  maximum  stress  in  each  member  directly  by 
one  equation,  as  we  have  generally  done  ;  and  this  is  the 
simplest  method. 

Dead  panel  load  per  truss  =  6  tons. 

Train  panel  load  per  truss  =  12  tons. 

Locomotive  panel  load  per  truss  =  30  tons. 

Locomotive  excess  panel  load  per  truss  =  18  tons. 
tan  0  =  1;  sec  0  =  1.414. 

For  the  maximum  chord  stresses  (except  2-4)  the  loco- 
motive must  stand  at  4,  and  the  train  panel  loads  at  all  the 
other  joints  ;  for  the  maximum  stress  in  2-4  the  locomotive 
must  be  put  at  6,  as  can  easily  be  seen.*  Then, 

Max.  stress  in  2-4 

=  2  x  6  +  ^  x  30  +  if  (2  +  4  +  6)  =  50.4  tons. 

Max.  stress  in  4-6 

=  2  x  18  +  4  x  18  +  (ft  -  ^18  =  122.4  tons. 


*  This  method  does  not  give  strictly  the  maximum  stresses  in  all  the 
chord  members.  While  the  error  near  the  ends  of  the  truss  is  quite  small, 
it  increases  towards  the  middle. 


122 


ROOFS  AND  BRIDGES. 


The  minimum  chord  stresses  are  due  to  dead  load  alone. 
For  the  maximum  stress  in  any  diagonal  the  live  load 
is  placed  on  the  right  of  a  section  cutting  that  diagonal,  as 
by  the  usual  method.     Thus, 
Max.  stress  in  5-8 

=  [ii.  x  6  +  30  x  Tfr  + 1.2(1  +3  +  5)]  1.414  =  57.7  tons. 
Min.  stress  in  7-10 

=  [1  x  6  -  30  x  fV]  1.414  =  0.0. 

That  is,  the  dead  load  passing  through  7-10  to  the  left 
abutment  is  neutralized  by  the  part  of  the  locomotive  load 
passing  through  7-10  to  the  right  abutment. 

The  following  stresses  are  found  in  a  manner  similar  to 
the  above: 

UPPER  CHORD  STRESSES. 


MEMBERS. 

1-3 

3-5 

5-7 

7-9 

9-11 

Max  
Min  

-61.2 
-15.0 

-133.2 
—  390 

-183.6 
—  57.0 

-219.6 
—  69.0 

-234.0 
-  75.0 

LOWER  CHORD  STRESSES. 


MEMBERS. 

2-1 

4-6 

6-6 

8-10 

10-12 

Max  
Min  

+50.4 
+  12.0 

+  122.4 
+  36.0 

+  176.4 
+  54.0 

+208.8 
+  66.0 

+  226.8 
+  72.0 

DIAGONAL  STRESSES. 


MEMBERS. 

1-4 

3-6 

5-8 

7-10 

9-12 

11-10' 

Max     .         .  . 

+  86  5 

+  71  3 

+  57  7 

+44  1 

+  32  2 

+  20  3 

Min  

+21.2 

+  17.0 

+  8.6 

+  0.0 

-10.2 

-20.4 

Max.  stress  in  1-2  =  -  61.2  tons. 


BRIDGE   TRUSSES.  123 

Prob.  99.  A  deck  Pratt  truss,  Fig.  50,  has  10  panels,  each 
12|  feet  long  and  12£  feet  deep  ;  the  dead  load  is  800  Ibs. 

1        3        5        7       9       11        9'      T       5'      3'       1' 

KNKNXIXIXI/I/M 

2  A       4       6        8       10      12       10'     8'      6'      U'      "  2' 
Fig.  SO 

per  foot  per  truss,  and  the  train  load  is  1600  Ibs.  per  foot 
per  truss  preceded  by  two  locomotive  panel  loads  of  30  tons 
each  per  truss  :  find  the  stresses  in  all  the  members. 

Consider  three  fifths  of  the  dead  load  as  applied  at  the 
upper  chord,  and  two  fifths  at  the  lower  chord.* 

Dead  panel  load  per  truss  =  5  tons. 
Train  panel  load  per  truss  =  10  tons. 
Excess  panel  load  per  truss  =  20  tons. 

For  maximum  chord  stresses  put  locomotive  panel  loads 
at  joints  3  and  5,  and  the  train  panel  loads  at  all  the  other 
joints.  Thus, 

Max.  stress  in  1-3 

=  41  x  15  +  20  (^  +  J^)  =  101.5  tons  =  stress  in  4-6. 

Max.  stress  in  3-5 


Max.  stress  in  5-7 

=  101  x  15  +  ffl  (17  +  7  -  3)  =  199.5  tons. 

For  maximum  stress  in  any  diagonal  the  live  load  is 
placed  on  the  right,  by  the  usual  method. 

*  In  railroad  bridges  it  is  customary  to  take  two  thirds  of  the  dead  load 
as  applied  at  the  loaded  chord  ;  that  is,  the  chord  which  carries  the  live 
load,  and  one  third  at  the  unloaded  chord. 


124 


EOOFS  AND  BRIDGES. 
CHORD  STRESSES. 


MEMBERS. 

1-3 

3-5 

5-7 

7-9 

9-11 

Max 

101  5 

168  0 

199  5 

216  0 

217  5 

Min  

-  22.5 

-  40.0 

-  52.5 

-  60.0 

-  62.5 

STRESSES  IN  THE  DIAGONALS. 


MEMBERS. 

1-4 

3-6 

5-8 

7-10 

9-12 

8-9 

10-11 

Max..  . 

Min.  .  . 

+  143.5 
+  31.8 

+  118.0 
+  20.5 

+94.0 
+  4.9 

+  71.3 
0.0 

+50.2 
0.0 

+  12.1 

0.0 

+30.4 
0.0 

Max.  compression  in  the  posts=103.0,  99.5,  81.5,  64.5,  48.5,  33.5  tons. 

Prob.  100.  A  through  Howe  truss,  Fig.  32,  has  10  panels, 
each  12  feet  long  and  12  feet  deep;  the  dead  load  is  1000 
Ibs.  per  foot  per  truss,  and  the  train  load  is  1667  Ibs.  per 
foot  per  truss,  preceded  by  two  locomotive  panel  loads  of 
30  tons  each  per  truss :  find  the  stresses  in  all  the  members. 

Dead  panel  load  =  6  tons. 
Train  panel  load  =  10  tons. 
Excess  panel  load  =  20  tons. 

Consider  one  third  of  the  dead  load  as  applied  at  the 
upper  chord. 

CHORD  STRESSES. 


MEMBERS. 

2-4 

4-6 

6-8 

8-10 

10-12 

Max  

+  106.0 

+  176.0 

+  210.0 

+  228.0 

+230.0 

Min 

+  27  0 

+  48  0 

+  63.0 

+  72.0 

+  75.0 

BRIDGE  TRUSSES. 
STRESSES  IN  THE  DIAGONALS. 


125 


MEMBERS. 

2-3 

4-5 

6-7 

8-9 

10-11 

7-10 

9-12 

Max.    .  . 

-149.9 

-123.0 

-97.6 

-73.6 

-50.9 

-9.9 

-29.7 

Min.     .  . 

-  38.0 

-  25.5 

-  8.5 

0.0 

0.0 

0.0 

0.0 

STRESSES  IN  THE  VERTICALS. 


MEMBERS. 

3-4 

5-6 

7-S 

9-10 

11-12 

Max               

+  104.0 

+  85.0 

+  67  0 

+  500 

+34  0 

Min  

+  25.0 

+  16.0 

+  4.0 

+  4.0 

+  40 

Prob.  101.  A  through  Pratt  truss,  Fig.  33,  has  10  panels, 
each  15  feet  long  and  15  feet  deep ;  the  dead  load  is  1200 
Ibs.  per  foot  per  truss,  and  the  train  load  is  2000  Ibs.  per  foot 
per  truss,  preceded  by  two  locomotive  panel  loads  of  30  tons 
each  per  truss :  find  the  stresses  in  all  the  members. 

Consider  one  third  of  the  dead  load  as  applied  at  the 
upper  chord. 

CHORD  STRESSES. 


MEMBERS. 

2-4 

4-6 

6-8 

8-10 

10-12 

9-11 

Max.  .  . 
Min.  .  . 

+  133.5 
+  40.5 

+  133.5 
+  40.5 

+228.0 
+   72.0 

+283.5 
+  94.5 

+  315.0 
+  108.0 

-322.5 
-112.5 

STRESSES  IN  THE  DIAGONALS. 


MEMBERS. 

M 

3-0 

E-S 

7-10 

8-12 

8-9 

1C-11 

Max  

-188.8 

+  152.7 

+  118.8 

+87.0 

+  57.3 

+4.2 

+29.7 

Min  

-  57.3 

+   40.3 

+  19.1 

0.0 

0-0 

0.0 

0.0 

126 


ROOFS  AND  BRIDGES. 
STRESSES  IN  THE  VERTICALS. 


MEMBERS. 

3-4 

£-8 

7-3 

9-10 

11-12 

Max.    

+  36.0 

—87  0 

-64  5 

-43  5 

-24.0 

Min. 

+  60 

16  5 

3  0 

3  0 

3  0 

Prob.  102.  A  through  Whipple  truss,  Fig.  37,  has  12 
panels,  each  12  feet  long  and  24  feet  deep ;  the  dead  load 
per  foot  per  truss  is  1000  Ibs.,  and  the  train  load  is  2000  Ibs. 
per  foot  per  truss,  preceded  by  one  locomotive  panel  load  of 
30  tons  per  truss :  find  the  stresses  in  all  the  members. 

For  max.  chord  stresses  put  locomotive  panel  load  at  joint 
4  and  the  train  panel  loads  at  all  the  other  joints.  * 

Dead  panel  load  =  6  tons. 
Train  panel  load  =  12  tons. 
Excess  panel  load  =  18  tons. 
Max.  stress  in  4-6  =  (3  x  18  +  ||  x  18)  x  .5 

=  35.3  tons. 

Max.  stress  in  6-8    =  35.3  +  2|  x  18  =  80.3  tons. 
Max.  stress  in  8-10  =  80.3  +  2  x  18  -  ^  x  18 

=  114.8  tons. 

CHORD  STRESSES. 


MEMBERS. 

2-4 

4-6 

6-S 

8-10 

10-12 

12-H 

9-11 

Max.  .  .  . 
Min.  .  .  . 

0.0 
0.0 

+35.3 
+  9.0 

+  80.3 
+  24.0 

+  114.8 
+  36.0 

+  141.8 
+  45.0 

+  158.3 
+  51.0 

-167.3 
-   54.0 

*  Put  all  the  dead  load  on  the  loaded  chord,  unless  otherwise  stated. 


BRIDGE  TRUSSES. 


127 


STRESSES  IN  THE  DIAGONALS. 


MEMBERS. 

1-t 

1-6 

3-3 

5-10 

7-12 

9-14 

11-12' 

18-10 

ll-S 

Max.  .  . 

+  78.8 

+  84.8 

+  71.4 

+58.0 

+46.0 

+33.9 

+23.3 

+  12.7 

+3.5 

Min.    .  . 

+20.1 

+21.2 

+  13.4 

+  5.7 

0.0 

0.0 

0.0 

0.0 

0.0 

STRESSES  IN  THE  VERTICALS. 


MEMBERS. 

1-2 

S-4 

5-6 

T-S 

9-10 

11-12 

13-14 

Max.     .  . 

-115.5 

-50.5 

-41.0 

-32.5 

-24.0 

-16.5 

-9.0 

Min.     .  . 

-  33.0 

-  9.5 

-  4.0 

0.0 

0.0 

0.0 

0.0 

Prob.  103.  A  through  Whipple  truss,  Fig.  38,  has  16 
panels,  each  10  feet  long  and  20  feet  deep ;  the  dead  and 
train  loads  are  1000  Ibs.  and  3000  Ibs.  per  foot  per  truss, 
the  train  being  preceded  by  one  locomotive  panel  load  of  30 
tons  per  truss :  find  the  stresses  in  all  the  members. 

Art.  41.  Method  of  Calculating  Stresses  when  one 
Concentrated  Excess  Load  accompanies  a  Uniform 
Train  Load.  —  This  method  is  sometimes  used,  like  the 
one  in  Art.  40,  to  avoid  the  practice  of  finding  the  stresses 
due  to  the  locomotive  wheel  loads.  But  we  have  seen  that 
the  method  in  Art.  40  does  not  give  strictly  the  maximum 
stresses  in  all  the  chord  members. 

Let  Fig.  51  be  a  truss  supporting  a  uniform  train  load 
covering  the  span,  and  a  concentrated  excess  load  P.  We 
suppose  the  excess  load  P  to  be  the  difference  between  the 
locomotive  panel  load  and  the  uniform  train  panel  load  as 
in  Art.  40.  Then  the  stress  in  any  chord  member,  as  bd, 
caused  by  the  concentrated  load  P,  is  equal  to  the  bending 
moment  at  c,  the  center  of  moments  for  bd,  divided  by  the 
lever  arm  for  the  chord ;  and  hence  the  chord  stress  will  be 


128  ROOFS  AND   BRIDGES. 

a  maximum  when  the  concentrated  load  P  is  so  placed  as  to 
make  the  bending  moment  a  maximum.  Now,  for  a  single 
concentrated  load,  the  maximum  bending  moment  at  any 

h      (7>p 


point  occurs  when  the  load  is  at  that  point ;  for,  if  the  load 
be  moved  to  either  side  of  the  point,  the  reaction  of  the 
opposite  abutment  will  be  diminished,  and  hence  the  mo- 
ment will  be  diminished. 

Therefore,  for  a  concentrated  excess  load  and  a  uniform 
train  load,  the  maximum  bending  moment  at  any  point,  and 
consequently  the  maximum  chord  stress  in  any  member,  occurs 
when  the  concentrated  load  is  at  the  center  of  moments  for 
that  member,  or  at  the  vertical  section  through  the  center  of 
moments,  and  the  uniform  train  load  covers  the  whole  span 
(Art.  22). 

Thus,  for  the  maximum  stress  in  bd,  the  concentrated 
load  P  is  at  c,  the  center  of  moments  for  bd ;  and  so  for  any 
other  panel  in  the  lower  chord,  or  unloaded  chord  generally. 
For  any  panel  in  the  upper,  or  loaded  chord,  as  he,  of  a 
truss  like  the  Warren,  the  concentrated  load  P  acts  at  c, 
or  at  that  end  of  the  panel  which  is  to  the  right  of  the  ver- 
tical section  through  b,  the  center  of  moments  for  he,  while 
the  uniform  train  load  covers  the  whole  span  (Art.  22).  For 
any  panel  in  the  loaded  chord  of  a  truss  like  the  Pratt  or 
Howe,  where  the  apexes  of  the  upper  chord  are  vertically 
above  those  of  the  lower  chord,  the  concentrated  load 
P  is  at  the  apex  directly  over  or  under  the  center  of  mo- 
ments for  that  member. 


BRIDGE  TRUSSES.  129 

For  a  single  concentrated  load,  the  maximum  positive 
shear  at  any  section  will  occur  when  the  load  is  just  to 
the  right  of  the  section;  for  the  left  reaction  is  then  a 
maximum. 

TJierefore,  for  a  concentrated  excess  load  and  a  uniform 
train  load,  the  maximum  stress  in  any  brace  occurs  when  the 
concentrated  load  is  at  the  panel  point  immediately  on  the 
riyht  of  the  section,  and  the  uniform  train  load  covers  the  span 
from  the  right  abutment  to  this  same  panel  point. 

Thus,  the  greatest  positive  shear,  and  therefore  the 
maximum  stress,  in  be  or  in  bh,  occurs  when  the  con- 
centrated load  is  at  the  apex  c  and  the  uniform  train  load 
extends  from  this  apex  to  the  right  abutment.  The 
greatest  negative  shear  for  be  or  for  bh  would  occur 
when  the  concentrated  load  is  at  the  apex  h  and  the  uni- 
form train  load  reaches  from  this  apex  to  the  left  abutment. 
For  maximum  chord  stresses,  cars  both  precede  and  follow 
the  locomotive.  This  does  not  often  happen.  For  maxi- 
mum stresses  in  the  braces  the  locomotive  precedes  the 
cars.  It  follows  therefore,  that  maximum  chord  stresses 
are  of  less  frequent  occurrence  than  maximum  web  stresses, 
which  occur  for  every  passage  of  the  train. 

Prob.  104.  A  through  Howe  truss,  Fig.  32,  has  10  panels, 
each  12  feet  long  and  12  feet  deep;  the  dead  and  train 
loads  are  1000  Ibs.  and  2000  Ibs.  per  foot  per  truss,  and  the 
locomotive  panel  load  is  30  tons  per  truss:  find  the  stresses 
in  all  the  members. 

Consider  one  third  of  the  dead  load  as  applied  at  the 
upper  chord. 

Dead  panel  load    =    6  tons. 

Train  panel  load    =12  tons. 

Excess  panel  load  =  18  tons. 


130 


ROOFS  AND  BRIDGES. 


For  the  maximum  stress  in  any  chord  member,  as  6-8, 
the  excess  load  is  placed  at  the  joint  8  and  the  train  load 
covers  the  whole  span.     Then,  (2)  of  Art.  19, 
Max.  stress  in 

6-8  =  10.5  xl8  +  i^x!8x3=  226.8  tons. 
Otherwise  by  moments,  (1)  of  Art.  19,  thus: 
Left  reaction  =  4|  x  18  +  -^  x  18  =  93.6  tons. 
The  equation  of  moments  about  the  point  7  is 

93.6  x  36  -  18(12  +  24)  -  stress  in  6-8  x  12  =  0. 
.-.  max.  stress  in 

6-8  =  93.6  x  3  -  18  x  3  =  226.8  tons,  as  before. 
For  the  maximum   stress   in   any  diagonal,  as   4-5,  the 
excess  load  is  placed  at  6,  and  the  train  loads  cover   the 
span  from  this  point  to  the  right  abutment.     Then, 
Max.  stress  in 
4_5  =  _  [3£  x  6  +  1.2(1  +  2  +  -  +  8)  +  .8  x  18]1.414 

=  - 111.1  tons. 
Min.  stress  in 
6-7  =  -  [21  x  6  -  if  x  3  -  ^  x  18]  1.414  =  -  11  tons. 

CHORD   STRESSES. 


MEMBERS. 

2-4 

4-6 

6-8 

8-10 

10-12 

Max 

+  97.2 

-)-  172.8 

+  226  8 

+  2592 

+  2700 

Min  

-f  27.0 

+    48.0 

+    63.0 

+    72.0 

+    75.0 

STRESSES   IN   THE    DIAGONALS. 


MEMBERS. 

2-3 

4-5 

e-7 

8-9 

10-11 

7-10 

9-12 

JUax.    .  . 

-137.4 

-111.1 

-86.5 

-63.6 

-42.4 

-5.1 

-22.9 

Min.     .  . 

-  38.2 

-  25.4 

-11.0 

0.0 

0.0 

0.0 

0.0 

BRIDGE  TRUSSES. 
STRESSES  IN  THE  VERTICALS. 


131 


MEMBERS. 

3-t 

5-6 

7-8 

9-10 

]•_•._> 

Max  

+95.2 

+  76.6 

+  59.2 

+43.0 

+34  0 

Min  

+  25.0 

+  16.0 

+  5.8 

+  4.0 

+  4.0 

Prob.  105.  A  through  Pratt  truss,  Fig.  33,  has  10  panels, 
each  15  feet  long  and  15  feet  deep ;  the  dead  and  train  loads 
are  1200  Ibs.  and  2000  Ibs.  per  foot  per  truss,  and  the  excess 
load  is  20  tons  per  truss  :  find  the  stresses  in  all  the  members. 
Consider  one  third  of  the  dead  load  as  applied  at  the  upper 
chord. 

Dead  panel  load  =  9  tons. 
Train  panel  load  =  15  tons. 
Excess  panel  load  =  20  tons. 

Max.  stress  in  8-10  =  10  £  x  24  +•  &  x  20  x  3  =  294  tons. 
Max.  stress  in 

3-6  =  [31  x  9  +  if  x  36  +  f£  x  8]  1.414  =  143.5  tons. 
Max.  stress  in 
7-8  =  -  [li  x  9  +  3  + 1.5  x  21  +  2  x  6]  =  -  60  tons. 

CHORD  STRESSES. 


MEMBERS. 

2-4 

4-6 

6-S 

8-10 

10-12 

9-11 

Max.  .  . 
Min.  .  . 

+  126.0 
+  40.5 

+  126.0 
+  40.5 

+  224.0 
+  72.0 

+294.0 
+  94.5 

+336.0 
+  108.0 

-350.0 
-112.5 

STRESSES  IN  THE  DIAGONALS. 


MEMBERS. 

2-3 

3-6 

5-8 

7-10 

9-12 

8-9 

10-11 

Max.    . 

-178.2 

+  143.6 

+  111.0 

+  80.6 

+52.3 

+2.1 

+26.2 

Min.     . 

-  57.3 

+  39.6 

+  19.8 

0.0 

0.0 

0.0 

0.0 

132  ROOFS  AND  BRIDGES. 

STRESSES  IN  THE  VERTICALS. 


MEMBERS. 

3-4 

5-6 

7-S 

9-10 

11-12 

Max  

41.0 

—81.5 

-60  0 

-40.0 

-26.0 

Min 

-f-  6  0 

17  0 

3  0 

3  0 

3  0 

Prob.  106.  A  double  Warren  truss,  Fig.  49,  used  as  a 
deck  bridge,  has  10  panels,  each  14  feet  long  and  14  feet 
deep ;  the  dead  and  train  loads  are  1000  Ibs.  and  2000  Ibs. 
per  foot  per  truss,  and  the  excess  load  is  20  tons  per  truss : 
find  the  stresses  in  all  the  members. 

UPPER  CHORD  STRESSES. 


MEMBERS. 

1-3 

3-5 

5-7 

7-9 

9-11 

Max.      . 

—68  0 

-154.0 

-225  0 

-271  0 

-292  0 

Min  

-14.0 

-  42.0 

—  63.0 

—  77.0 

-  84.0 

LOWER  CHORD  STRESSES. 


MEMBERS. 

2-i 

4-6 

6-3 

8-10 

10-12 

Max  
Min  

+  70.5 
+  17.5 

+  168.5 
+  45.5 

+  241.5 
+  66.5 

+  289.5 
+  80.5 

+312.5 
+  87.5 

DIAGONAL  STRESSES. 


MEMBERS. 

1-4 

3-6 

5-8 

7-10 

9-12 

11-10' 

Max     .  . 

+82  0 

+66  3 

+50  6 

+  36  9 

+  23  2 

+  11  5 

Min  

+  19.8 

+  10.0 

+  0.3 

-11.5 

-23.2 

-36.9 

Max.  stress  in  1-2  =  2    x  21  + 10  =  62.5  tons. 


BRIDGE   TRUSSES.  133 

Prob.  107.  A  deck  Pratt  truss,  Fig.  50,  has  10  panels, 
each  15  feet  long  and  15  feet  deep ;  the  dead  and  train  loads 
are  800  Ibs.  and  2000  Ibs.  per  foot  per  truss,  and  the  excess 
load  is  20  tons  per  truss:  find  the  stresses  in  all  the 
members. 

Art.  42.  Method  of  Calculating  Stresses  when  two 
equal  Concentrated  Excess  Loads,  placed  about  50 
feet  apart,  accompany  a  Uniform  Train  Load.  —  This 
method  is  often  used,  like  the  two  in  Arts.  40  and  41,  to 
avoid  the  practice  of  finding  the  stresses  due  to  the  loco- 
motive wheel  loads,  and  is  a  nearer  approximation  to  the 
actual  loads  than  either  of  the  other  two. 

Let  Fig.  52  be  a  truss  supporting  a  uniform  train  load 
covering  the  span,  and  two  equal  concentrated  excess  loads, 


Pl  and  P2.  As  in  Arts.  40  and  41,  we  suppose  each  excess 
load  to  be  the  difference  between  the  locomotive  panel  load 
and  the  uniform  train  panel  load.  Then  the  stress  in  any 
chord  member,  as  bd,  caused  by  the  concentrated  loads  PI 
and  P2  is  equal  to  the  bending  moment  at  c,  the  center  of 
moments  for  bd,  divided  by  the  lever  arm  of  bd ;  and  hence 
the  chord  stress  will  be  a  maximum  when  the  two  con- 
centrated loads  are  so  placed  as  to  make  the  bending 
moment  a  maximum. 

Now,  for  two  equal  loads,  a  fixed  distance  apart,  the 
maximum  moment  at  any  point  occurs  when  one  load  is  at 


134  ROOFS.  AND  BRILGES. 

the  point  and  the  other  is  on  the  longer  segment  of  the 
truss.     This  may  be  shown  as  follows  : 

Let  I  =  length  of  span  in  Fig.  52,  d  =  distance  between 
the  two  equal  loads  Pl  and  P2,  and  a  =  distance  from  left 
abutment  to  center  of  moments  c;  then  supposing  the  load 
PI  to  be  at  a  distance  x  to  the  left  of  c,  and  calling  M  the 
moment  at  the  point  c,  due  to  the  two  loads,  P^  and  P2,  and 
denoting  each  load  by  P,  since  they  are  equal,  we  have 

jy  _  p  a(2  Z  —  2  o  —  d)  —  (Z  —  2a)x 


maximmn) 


Similarly,  if  the  load  Px  be  placed  at  a  distance  x  to  the 
right  of  c,  the  moment  at  the  point  c  will  be  a  maximum 
when  x  =  0. 

Therefore,  for  two  equal  concentrated  excess  loads,  and  a 
uniform  train  load,  the  maximum  bending  moment  at  any 
point,  and  consequently  the  maximum  chord  stress  in  any 
member,  occurs  when  one,  of  the  equal  concentrated  loads  is  at 
the  center  of  moments  for  that  member,  and  the  other  con- 
centrated load  is  on  the  longer  segment  of  the  truss,  while  the 
uniform  train  load  covers  the  whole  span. 

Thus,  for  the  maximum   stress  in  bd,  the   concentrated 

load  P!  is  at  c,  and  the  other  load  P2  is  to  the  right  ;  and  so 

for  any  other  panel  in  the  lower,  or  unloaded,  chord  of  the 

left  half  of  the   truss.     For  any  panel   in  the   upper,  or 

*  loaded,  chord  of  the  left  half,  as  he,  of  a  truss  like  the 

/   Warren,  where  all  the  members  are  inclined,  the  load  PI  acts 

at  c,  or  at  that  end  of  the  panel  which  is  to  the  right  of  the 


BRIDGE   TRUSSES.  135 

vertical  section  through  b,  the  center  of  moments  for  he,  and 
the  other  load  P2  is  to  the  right. 

For  any  panel  in  the  loaded  chord  of  a  truss  like  the 
Pratt  or  Howe,  where  the  upper  apexes  are  vertically  above 
the  corresponding  lower  ones,  the  concentrated  load  P±  is  at 
the  apex  directly  over  or  under  the  center  of  moments  for 
that  member. 

For  two  equal  concentrated  loads  Pl  and  Pz,  the  maximum 
positive  shear  at  any  section  occurs  when  both  loads  are  on 
the  right  of  the  section  and  Pl  is  as  near  to  it  as  possible ; 
for  the  left  reaction  is  then  a  maximum. 

Tlierefore,  for  two  equal  concentrated  excess  loads  and  a, 
uniform  train  load,  the  maximum  stress  in  any  brace  o<xurs 
when  both  loads  are  on  the  right  of  the  section  and  one  of  them 
is  at  the  panel  point  immediately  on  the  right  of  the  section, 
and  the  uniform  train  load  covers  the  span  from  this  point  to 
the  right  abutment. 

The  second  excess  load  P2  should  be  placed  at  a  panel 
point  at  an  interval  of  about  50  feet  to  the  right  of  P»  to 
simplify  the  computation.  Thus,  should  the  panel  length 
be  12, 15,  or  18  feet,  the  distance  between  the  loads  would 
be  48,  45,  or  54  feet. 

Prob.  108.  A  through  Warren  truss,  Fig.  53,  has  8  panels, 
each  10  feet  long  and  10  feet  deep,  its  web  members  all 


forming  isosceles  triangles;  the  dead  and  train  loads  are 
1000  Ibs.  and  2000  Ibs.  per  foot  per  truss,  and  there  are  two 


136 


ROOFS  AND  BRIDGES. 


excess  loads  50  feet  apart,  each  of  33  tons :  find  the  stresses 
in  all  the  members. 

Dead  panel  load  =    5  tons. 

Train  panel  load  =  10  tons. 

Each  excess  load  =  33  tons. 

tan0  =  i;  sec0  =  1.117. 

For  the  maximum  stress  in  any  chord  member,  as  4-6, 
one  excess  load  is  placed  at  6  and  the  other  one  at  4',  or  50 
feet  to  the  right  of  the  first,  and  the  dead  and  train  loads 
cover  the  whole  span.  Then  (2)  of  Art.  19, 

Max.  stress  in  4-6  =  [9|  x  15  +  -3/ (1  +  6) 3]  x  .5 

=  114.6  tons. 

Otherwise  by  moments,  (1)  of  Art.  19,  as  follows : 
Left  reaction  =  3£  x  15  +  33  (£  +  f)  =  81.375  tons. 
The  equation  of  moments  about  the  point  3  is 

81.375  x  15  —  15  x  5  —  stress  in  4-6  x  10  =  0. 
.-.  max.  stress  in  4-6  =  114.6  tons,  as  before. 
For  the  maxinmm  stress  in  any  diagonal,  as  3-6,  one 
excess  load  is  placed  at  6  and  the  other  50  feet  to  the  right 
of  it  at  4',  and  the  train  loads  cover  the  span  from  6  to  the 
right  abutment.     Then, 

Max.  stress  in  3-6  =  [2{-  x  5  +  -1/-  C1  +  2  -I h  6) 

+  sp  (6  +  1)  ]  1.117  =  75.5  tons  =  -  stress  in  3-4. 
Miu.  stress  in  5-6  =  -  [1|  x  5  -  f  x  10  -  f  x  33]  1.117. 
=  5.0  tons. 

UPPER  CHORD  STRESSES. 


MEMBERS. 

1-3 

3-5 

5-7 

7-7' 

Max 

89  6 

147  8 

174  4 

186  0 

Min                    

-17.5 

-  300 

-  37  5 

400 

BRIDGE  TRUSSES. 
LOWER  CHORD  STRESSES. 


137 


MEMBERS. 

2-1 

4-3 

... 

8-10 

Max                

+  44  8 

+  114.6 

+  152.8 

+  174.0 

Min           

+  8.8 

+  23.8 

+  33.8 

+  38.8 

DIAGONAL  STRESSES. 


MEMBERS. 

1-2 

1-4 

3-8 

5-3 

7-10 

Max  
Min  

-100.1 
-  19.5 

+  100.1 
+   19.5 

+  75.5 
+  8.0 

+52.4 
-  5.0 

+35.2 
-19.4 

Prob.  109.  A  deck  Pratt  truss,  Fig.  50,  has  10  panels, 
each  12  feet  long  and  12  feet  deep ;  the  dead  and  train  loads 
are  1000  Ibs.  and  2000  Ibs.  per  foot  per  truss,  and  there  are 
two  excess  loads  48  feet  apart,  each  of  30  tons :  find  the 
stresses  in  all  the  members. 

CHORD  STRESSES. 


MEMBERS. 

1-3 

8-5 

5-7 

7-9 

9-11 

Max.     .  .  . 
Min  

-123.0 
-  27.0 

-216.0 
-  48.0 

-279.0 
-  63.0 

-312.0 
-  72.0 

-315.0 
-  75.0 

STRESSES  IN  DIAGONALS. 


MEMBERS. 

1-4 

C-6 

C-1 

7-10 

9-12 

?-D 

10_11 

Max.    .  . 

+  173.9 

+  141.7 

+  111.1 

+82.3 

+55.1 

+  10.2 

+  2".' 

Min.     .  . 

+  38.2 

+  23.8 

+     7.6 

0.0 

0.0 

0.0 

(J.-> 

138 


EOOFS  AND   BRIDGES. 
STRESSES  IN  VERTICALS. 


MEMBERS. 

w 

£-1 

C-3 

7-8 

9-10 

11-12 

Max  
Min  

-138.0 
-  30.0 

-123.0 
-  27.0 

-100.2 
-  16.8 

-78.6 
-  6.0 

-58.2 
-  6.0 

-48.0 
-  6.0 

Prob.  110.  A  deck  Whipple  truss,  Fig.  37,  has  12  panels, 
each  12  feet  long  and  24  feet  deep ;  the  dead  and  train  loads 
are  1000  Ibs.  and  3000  Ibs.  per  foot  per  truss,  and  there  are 
two  excess  loads  48  feet  apart,  each  of  30  tons :  find  the 
stresses  in  all  the  members. 

Dead  panel  load  =  6  tons. 
Train  panel  load  =  18  tons. 
Each  excess  load  =  30  tons. 

tan  0  =  0.5;  tan0'  =  l;  sec  0  =  1.117;  sec  0' =  1.414. 
Max.  stress  in  4-6  =  [3x24+ff  (11+7)]  X. 5=58.5  tons. 
Max.  stress  in  6-8=4x24+2.5  (10+6)1^=136.0  tons. 
Max.  stress  in  3-8 

=  [2x6+1.5x25+2.5x14]  1.414=119.5  tons. 
Min.  stress  in  5-10= [9-1.5x2 -2.5x2]  1.414= 1.4  tons. 

CHORD  STRESSES. 


MEMBER?. 

2-i 

4-6 

6-3 

8-10 

10-12 

12-14 

9-11 

Max.  . 
Min.  . 

+00.0 
0.0 

+58.5 
+  9.0 

+  136.0 
+  24.0 

+  196.5 
+  36.0 

+240.0 
+  45.0 

+  266.5 
+  61.0 

-276.0 
-  54.0 

BRIDGE  TRUSSES.  139 

STRESSES  IN  DIAGONALS. 


MEMBERS. 

1-1 

1-6 

3-S 

6-10 

7-12 

9-14 

11-12' 

13-10 

11-3 

Max  

+130.7 

+141.4 

+  119.5 

+97.6 

+77.S 

+58.0 

+40.8 

+22.6 

+10.6 

Min  

+  20.1 

+  21.2 

+  11.8 

+  1.4 

0.0 

0.0 

0.0 

0.0 

0.0 

STRESSES  IN  VERTICALS. 


MBMBEBS. 

1-3 

8-1 

5-3 

7-3 

9-10 

11-12 

13-14 

Max.  .  . 

-179.0 

-117.0 

-100.0 

-84.5 

-69.0 

-55.0 

-54.0 

Min.    .  . 

-  36.0 

-  18.0 

-  15.0 

-  8.0 

-  6.0 

-  6.0 

-  6.0 

Prob.  111.  A  through  parabolic  bowstring  truss,  Fig.  44, 
has  8  panels,  each  10  feet  long,  and  10  feet  center  depth; 
the  verticals  are  ties,  and  the  diagonals  are  struts ;  the  dead 
and  train  loads  are  1000  Ibs.  and  3000  Ibs.  per  foot  per  truss, 
and  there  are  two  excess  loads  50  feet  apart,  each  of  30  tons : 
find  the  stresses  in  all  the  members. 

Dead  panel  load  =  5  tons. 
Train  panel  load  =  15  tons. 
Each  excess  load  =  30  tons. 

Length  of  3-4  =  4.375  feet ;  length  of  5-6  =  7.5  feet. 
•Length  of  7-8  =  9.375  feet;  length  of  9-10  =  10.0  feet. 

Then,  for  the  maximum  chord  stresses,  a  full  dead  and 
train  panel  load  must  be  at  each  lower  apex.  For  any 
member  of  the  lower  chord,  as  2-4,  one  excess  load  of  30 
tons  is  placed  at  4  and  the  other  at  6',  or  50  feet  to  the  right 
of  the  first,  and  the  dead  and  train  loads  cover  the  whole 
span,  as  just  stated.  Here  we  cannot  use  the  "method  of 
chord  increments,"  since  the  chords  are  not  parallel,  but 
must  use  the  "  method  of  moments,"  (1)  of  Art.  19. 


140  ROOFS  AND  BRIDGES. 

Eeaction  at  the  left  end  =  3£  x  20  +  30  ( £  +  f )  =  103.75 
tons. 

Max.  stress  in  2-4  = 103'75  x  10  =  237.1  tons. 
4.375 

This  tensile  stress  in  2-4  is  equal  to  the  horizontal  com- 
ponent of  the  compressive  stress  in  2-3,  by  (1)  of  Art.  5. 
Therefore  the  stress  in  2-3  is  equal  to  the  stress  in  2-4 
multiplied  by  the  secant  of  the  angle  between  2-3  and  2-4. 
Thus, 

Max.  stress  in  2-3  =  -  237.1  x  1.092  =  -  258.9  tons. 

For  the  maximum  stress  in  3-5,  one  excess  load  is  put 
at  6  and  the  other  at  4',  while  the  dead  and  train  loads 
cover  the  whole  span;  then  take  center  of  moments  at 
intersection  of  4-5  and  4-6  or  at  4.  And  so  on  for  the 
stresses  in  5-7  and  7-9. 

The  diagonal  stresses  are  found  by  putting  only  the  train 
and  excess  loads  on  the  truss  in  the  proper  position  for  each 
member,  since  the  dead  load  produces  no  stress  in  the  diag- 
onals (Art.  34).  Thus,  to  find  the  maximum  stress  in  4-5, 
one  excess  load  is  placed  at  6  and  the  other  50  feet  to  the 
right  of  it  at  4',  and  the  train  loads  cover  the  span  from  6  to 
the  right  abutment. 

Now  cut  3-5,  4-5,  and  4-6,  and  take  center  of  moments 
at  the  intersection  of  3-5  and  4-6,  which  is  4  feet  to 
the  left  of  2;  the  diagonal  3-6  is  not  in  action  for  this 
loading. 

Lever  arm  of  4-5  =  14  sin  0  =  14  x  ^  =  8.4  feet. 

12.5 
Left  reaction  for  this  loading 

=  1^(1  +  2  +  ...  +  6)  +  —(6  +  1)=  65.625  tons. 
8  8 

.-.  max.  stress  in  4-5  =  -  65'625  x  4  =  -  31.25  tons. 
8.4 


BRIDGE   TRUSSES. 


141 


UPPER  CHORD  STRESSES. 


MEMBERS. 

2-3 

8-5 

5-7 

7-9 

Max  

-258.8 

—230.5 

-213.7 

-208.3 

Min 

43  6 

41  0 

40  7 

40  1 

LOWER  CHORD  STRESSES. 


•MEMBERS. 

2-1 

4-6 

6-3 

8-10 

Max  

Min 

+237.1 
+  40  0 

+230.0 
+  40  0 

+220.0 
+  40  0 

+220.0 
+  40  0 

STRESSES  IN  THE  DIAGONALS. 


MEMBERS. 

4-5 

6-7 

8-9 

3-6 

6-8 

7-10 

Max  
Min  

-31.3 
0.0 

-34.2 
0.0 

-38.2 
0.0 

-49.0 
0.0 

-43.75 
0  0 

-41.1 
0  0 

Max.  stresses  in  the  verticals  =  +  50.0  tons. 
Min.  stresses  in  the  verticals  =  +  5  tons. 

Prob.  112.  A  through  circular  bowstring  truss,  Fig.  45, 
has  6  panels,  each  15  feet  long,  its  web  members  all  form- 
ing isosceles  triangles,  with  the  upper  apexes  on  the  cir- 
cumference of  a  circle  whose  radius  is  75  feet,  the  center 
depth  of  the  truss  being  14f  feet ;  the  dead  and  train  loads 
are  720  Ibs.  and  2176  Ibs.  per  foot  per  truss,  and  there  are 
two  excess  loads  45  feet  apart,  each  of  24  tons:  find  the 
stresses  in  all  the  members. 

Dead  panel  load  =  5.4  tons. 

Train  panel  load  =  16.32  tons. 

Each  excess  load  =  24  tons. 


142 


ROOFS  AND  BRIDGES. 
CHORD  STRESSES. 


MEMBERS. 

2-3 

3-5 

5-7 

7-7' 

2-4 

4-6 

6-8 

Max.  .  .  . 

-149.6 

-163.0 

-149.2 

-137.1 

+  125.0 

+  130.8 

+  125.4 

Min.  .  .  . 

-  24.7 

-  27.2 

-  25.6 

-  25.0 

+  20.6 

+  22.9 

+  23.7 

WEB  STRESSES. 


MEMBERS. 

3-t 

5-6 

7-3 

7'-6' 

5'-4' 

Max  
Min     

+  29.9 
+  4.9 

+33.4 
—  11.9 

+  34.8 
-15.1 

+37.9 
-139 

+43.2 
-  7  1 

Prdb.  113.  A  through  Pratt  truss,  Fig.  33,  has  10  panels, 
each  12  feet  long  and  12  feet  deep ;  the  dead  and  train  loads 
are  1000  Ibs.  and  3000  Ibs.  per  foot  per  truss,  and  there  are 
two  excess  loads  48  feet  apart,  each  of  30  tons:  find  the 
stresses  in  all  the  members. 

Art.  43.  The  Baltimore  Truss.  —  This  truss,  Fig.  54, 
or  some  modification  of  it,  is  now  used  very  generally  for 
long  spans  when  it  is  desired  to  avoid  long  panel  lengths. 
Each  large  panel  is  divided  into  two  smaller  ones  by  insert- 


T    f 


s'   h 


Fig.  54 


ing  half  ties  and  subvertical  struts,  or  half  struts  and  sub- 
vertical  ties,  according  as  it  is  a  deck  or  a  through  truss. 
I       In  the  deck  form,  Fig.  54,  the  subverticals  ai,  bj,  ck,  etc., 
are  strained  only  by  the  panel  loads  which  they  directly 


ROOF  TRUSSES.  143 

support;  the  dotted  diagonals  4 k,  61,  6'm,  4'n  are  counters, 
not  being  in  action  for  full  load. 

Prob.  114.  A  deck  Baltimore  truss,  Fig.  54,  has  16  panels, 
each  10  feet  long  and  20  feet  deep ;  all  the  verticals  are 
posts,  and  all  the  inclined  pieces  are  ties;  the  dead  and 
train  loads  are  800  Ibs.  and  1600  Ibs.  per  foot  per  truss,  and 
there  are  two  excess  loads  50  feet  apart,  each  of  30  tons : 
find  the  stresses  in  all  the  members. 

Dead  panel  load  =  4  tons. 
Train  panel  load  =  8  tons. 
Each  excess  load  =  30  tons. 

The  maximum  chord  stresses  occur  when  the  dead  and 
train  loads  act  at  every  upper  apex  (Art.  20),  and  the  excess 
loads  act  at  the  proper  apexes  (Art.  42).  Thus  for  3  6,  we 
have  30  tons  at  6  and  at  9,  50  feet  to  the  right  of  b.  The 
center  of  moments  is  at  4,  the  intersection  of  3-4  and  2-4. 
Hence,  the  left  reaction  for  dead  and  live  loads  is, 

R  =  71  x  12  +  f  |  (13  +  8)  =  129.375  tons. 
Calling  S  the  stress  in  3  &,  we  have 

129.375  x  40  - 12  (30  +  20)  +  8  X  20  =  0 ; 

.-.  S  =  stress  in  3  6  =  —  228.8  tons  =  stress  in  6  5. 

The  stresses  in  1  a  and  a  3,  3  &  and  b  5,  5  c  and  c  7,  7  d  and 
d  9,  etc.,  must  always  be  the  same,  since  the  posts  at,  bj,  ck, 
etc.,  are  perpendicular  to  the  chords,  and  cannot  therefore 
cause  any  stresses  in  them. 

For  5  c  and  c  7.  we  must  put  the  excess  loads  at  c  and  7', 
and  take  the  center  of  moments  at  6,  the  intersection  of  5  k 
and  4-6 ;  and  so  on. 


144  ROOFS  AND  BRIDGES. 

For  any  lower  chord  member,  as  4-6,  we  have  30  tons  at 
5  and  at  e,  50  feet  to  the  right  of  5.  The  center  of  moments 
is  at  5.  Thus, 

R  =  7|  x  12  +  ff-  (12  +  7)=  125.625  tons. 

.-.  125.625  x  40  -  12  (30  +  20  + 10)  -  S  x  20  =  0. 

.•.  S  =  stress  in  4-6  =  215.3  tons. 

It  is  evident  that  the  maximum  stress  in  each  subvertical, 
ai,  bj,  ck,  dl,  etc.,  is  equal  to  a  full  panel  load,  4  +  8  +  30 
=  42  tons  compression. 

It  is  also  evident  that  half  of  the  load  on  ai,  bj,  ck,  etc.,  is 
transmitted  to  3,  5,  7,  etc.,  through  the  half  ties  i3,  j5, 
k7,  etc. 

Therefore  the  maximum  tension  in  each  of  the  half  ties 

i 3,  j5,  k7,  etc.,  =4  +  8  +  30sec  e  =  21  x  1.414  =  29.7  tons.1 

2 

The  maximum  stress  in  the  upper  part  of  any  main  tie, 
as  5k,  occurs  when  the  excess  loads  act  at  c  and  at  7',  50 
feet  to  the  right  of  c,  and  the  train  loads  are  at  all  the 
apexes  on  the  right  of  5  k.  The  shear  in  5  k  for  this  load- 
ing is, 

31  x  4  +  f  £  (11  +  6)+TV(ll  +  10  +  ...  +  1)=  78.875. 
.-.  max.  stress  in  5k  =  78.875  x  1.414  =  111.5  tons. 

The  maximum  stress  in  any  main  vertical,  as  5-4,  is 
equal  to  the  greatest  shear  in  5-4 ;  and  this  greatest  shear 

1  At  the  center  of  the  truss  these  ties  must  act  as  counters.  Thus,  take 
the  tie  1-9,  put  the  excess  loads  at  3  and  d,  and  the  train  loads  at  all  the 
apexes  from  the  left  abutment  to  d  inclusive.  Then  the  greatest  shear 
in  the  section  cutting  d-9 ,  1-9,  and  1-8  is  —  28  875  tons.  As  1-8  is  not  in 
action  for  this  loading,  this  is  the  vertical  component  of  the  stress  in 
1-9. 

.-.  max.  stress  in  1-9  =  28.875  X  1.414  =  +  40.8  tons. 


BRIDGE  TRUSSES.  145 

in  5-^i  occurs  when  the  excess  loads  act  at  5  and  at  e,  50 
feet  to  the  right  of  5,  and  the  train  loads  are  at  all  the 
apexes  from  the  right  abutment  to  5  inclusive.  Then  the 
greatest  shear  in  5-4  is  that  which  is  due  to  this  loading, 
together  with  the  four  dead  loads  at  5,  c,  7,  and  d,  half  of 
the  dead  load  at  9,  and  half  of  the  dead  load  at  b,  which  is 
transmitted  to  5  through  the  half  tie,;' 5.  Hence, 

Max.  stress  in  5-4  = 

-[5x4  +  f$(12  +  7)+  A  (12  +  -  +  1)] 
=  -  94.6  tons. 

The  maximum  stress  in  the  lower  part  of  any  main  tie, 
as  J4,  is  equal  to  the  maximum  stress  in  4-5  multiplied 
by  the  secant  of  the  angle  which  the  tie  makes  with  the 
vertical.  Hence, 

Max.  stress  in./ 4  =  94.6  x  1.414  =  133.8  tons. 

The  maximum  stress  in  6-1  occurs  when  the  excess  loads 
are  at  a  and  7,  and  the  train  loads  are  at  all  the  joints  from 
the  left  abutment  to  7  inclusive.  Then  the  shear  in  the 
section  cutting  d-9,  1-9,  and  1-8  is  —21.625  tons,  which  is 
the  vertical  component  of  the  stress  in  1-9,  since  1-8  is  not 
in  action.  The  load  at  d  produces  in  1—9  a  negative  shear 
of  2  tons,  so  that  the  difference,  or  19.625,  must  come  from 
the  member  6-1. 

.:  max.  stress  in  6-1  =  -f- 19.625  x  1.414  =  +  27.7  tons. 

Otherwise  as  follows:  The  live  loads  going  to  the  right 
abutment  through  the  panel  6-8  =  23.63  tons.  The  dead 
loads  crossing  this  at  the  section  cutting  d-9, 1-9,  and  1-8  are 
%  of  load  at  9  +  -|  of  load  at  d  =  4  tons.  The  difference  is 
19.63  tons,  for  which  the  panel  6-8  must  be  counterbraced. 

Thus  are  found  the  following  maximum  stresses : 


146  ROOFS  AND  BRIDGES. 

MAXIMUM  CHORD  STRESSES. 


1-3 

3-5 

5-7 

7-9 

2-4 

4-6 

6-3 

-136.9 

-228.8 

-281.6 

-295.6 

+  127.1 

+  215.3 

+264.4 

MAXIMUM  STRESSES  IN  UPPER  PART  OF  MAIN  TIES  AND  HALF  TIES. 


1-4 

3^' 

5-k 

7-Z 

3-i 

H 

7-/t 

9-J 

+  193.6 

+  151.1 

+  111.5 

+  74.8 

+29.7 

+29.7 

+29.7 

+  40.8 

MAXIMUM  STRESSES  IN  LOWER  PART  OF  MAIN  TIES. 


i-2 

J-* 

£-6 

Z-8 

6-Z 

4-Jfc 

+  174.8 

+  133.8 

+95.6 

+  60.3 

+27.7 

+0.7 

MAXIMUM  STRESSES  IN  THE  VERTICALS. 


2-3 

4-5 

6-7 

8-9 

ai,  ty,  ck,  dl. 

-123.6 

-94.6 

-67.6 

-59.6 

-42.0 

NOTE.  —  In  the  above  determination  of  the  maximum  stresses  in 
the  long  verticals  and  in  the  lower  ends  of  the  long  diagonals,  as.  for 
example,  in  5-4  and  j-4,  it  was  assumed  that  the  largest  possible  shear 
in  5-4  or  in  j-A  occurs  when  the  excess  loads  act  at  5  and  at  e,  and 
the  train  loads  are  at  all  the  apexes  from  the  right  abutment  to  the 
joint  5  inclusive.  It  is  evident,  however,  from  a  little  inspection  of 
Fig.  64,  that,  if  a  train  panel  load  be  placed  at  the  joint  ft,  one  half  of 
this  train  load  at  b  will  be  transmitted  to  the  joint  5  through  the  half 
tie  j  5,  just  as  one  half  of  the  dead  load  at  b  is  transmitted  to  5  through 
the  half  tie  j  5,  and  that  this  half  train  load  transmitted  to  5,  minus 


BRIDGE   TRUSSES. 


147 


the  part  that  is  transmitted  to  the  right  abutment,  which  is  ^  of  the 
train  load  at  6,  will  form  part  of  the  shear  in  5-4  or  in  j-4  ;  that  is, 
the  shear  will  be  increased  by  \  x  8  -  T%  x  8  =  ^  x  8  =  2.5.  Hence, 

Max.  stress  in  5-4  =  -  [5  x  4  +  ff  (12  +  7)  +  A(12  +  "*  +  *)  +  2-5] 
=  -97.1  tons. 

Max.  stress  in  j-4  =  97.1  x  1.414  =  137.3  tons. 

Similarly, 

Max.  stress  in  2-3  =  -  [7  x  12  +  f$(14  +  9)]  =  -  127.1  tons. 

Max.  stress  in  i-2  =  127.1  x  1.414  =  179.7  tons, 
and  so  on  for  6-7  and  Jk-6,  8-9,  and  1-8. 

Thus,  stress  in  6-7  =-68.6,  in  k-Q  =  97.0,  in  8-9  =-43.1,  in 
2-8  =  61  tons. 

It  is  evident  that  by  putting  a  train  panel  load  on  the  apex  just  to 
the  left  of  the  section  considered,  the  shear  in  that  section  is  increased 
the  most  when  near  the  left  end  of  the  truss ;  and  that  it  decreases  on 
approaching  the  middle  of  the  truss  without  becoming  zero. 

•  Prob.  115.  A  deck  Baltimore  truss,  Fig.  55,  has  14  panels, 
each  10  feet  long  and  20  feet  deep,  the  verticals  being  posts 
and  the  diagonals  ties ;  the  dead  and  train  loads  are  1000  Ibs. 


and  2000  Ibs.  per  foot  per  truss,  and  there  are  two  excess 
loads  50  feet  apart,  each  of  30  tons:  find  the  maximum 
stresses  in  all  the  members.  (See  note  to  last  problem.) 


MAXIMUM  CHORD  STRESSES. 


1-3 

3-5 

5-7 

7-7' 

2-4 

4-6 

6-C' 

-142.5 

-230.4 

-271.1 

-264.7 

+  130.7 

+214.3 

+260.7 

148  ROOFS  AND  BRIDGES. 

MAXIMUM  STRESSES  IN  UPPER  PART  OF  MAIN  TIES  AND  HALF  TIES. 


l-i 

H 

5-£ 

7-; 

8-i 

w 

1-k 

+201.5 

+  150.0 

+  102.5 

+59.1 

+  31.8 

+  31.8 

+  31.8 

MAXIMUM  STRESSES  IN  LOWER  PART  OF  MAIN  TIES. 


i-2 

J-* 

/fc-6 

c-l 

4-/fc 

+  184.8 

+  133.3 

+  85.8 

+42.4 

+  8.1 

MAXIMUM  STRESSES  IN  THE  VERTICALS. 


2-8 

4-5 

6-7 

ai,  ty,  ck,  dl. 

-130.7 

-94.3 

-60.7 

-45.0 

Prob.  116.  A  through  Baltimore  truss,  Fig.  56,  has  16 
panels,  each  16  feet  long  and  32  feet  deep ;  the  dead  load 
is  given  by  formula  (3),  Art.  15,  the  train  load  is  1500  Ibs. 

9  7'  5'  S' 


Fig.  SO 

per  foot  per  truss,  and  there  are  two  excess  loads  48  feet 
apart,  each  of  30  tons  :  find  the  maximum  stresses  in  all 
the  members. 


Dead  panel  load    = 

Train  panel  load   =  12  tons. 
Each  excess  load  =  30  tons. 


750 


16  =  g 


BRIDGE  TRUSSES.  149 

The  maximum  chord  stresses  occur  when  the  dead  and 
train  loads  act  at  every  lower  apex  (Art.  20),  and  the  excess 
loads  act  at  the  proper  apexes  (Art.  42).  Thus  for  66,  we 
have  30  tons  at  4  and  at  c,  48  feet  to  the  right  of  6.  The 
center  of  moments  is  at  3,  the  intersection  of  3-5  and  3-6. 
Hence  the  left  reaction  for  dead  and  live  loads  is, 
R  =  7\  x  20  +  f^  (14  +  11)  =  196.875  tons. 

Calling  s  the  max.  stress  in  b  6,  we  have, 

196.875  x  32  -  20  x  16  +  20  x  16  -  s  x  32  =  0. 

.-.  s  =  max.  stress  in  &6  =  196.9  tons  =  max.  stress  in  46. 

For  any  upper  chord  member,  as  5-7,  we  have  30  tons  at 
8  and  at  e,  48  feet  to  the  right  of  8.  The  center  of  moments 
is  at  8.  Thus, 

E  =  7£  x  20  +  f £  (10  +  7)  =  181.875  tons. 

.-.  181.875  x6x!6-20(16+32  +  48  +  64+80)  +  sx32  =  0. 
.-.  s  =  max.  stress  in  5-7  =  —  395.6  tons. 

Similarly,  the  other  chord  stresses  are  found. 

It  is  evident  that  the  maximum  stress  in  each  subvertical, 
01,  bJc,  d,  dm,  etc.,  is  equal  to  a  full  panel  load  8+12  +  30 
=  50  tons  tension. 

It  is  also  evident  that  half  of  the  load  on  ai,  bk,  cl,  dm, 
etc.,  is  transmitted  to  4,  4,  6,  8,  etc.,  through  the  half  struts 
t4,  A:  4,  Z6,  ra8,  etc. 

Therefore  the  maximum  compression  in  each  of  the  half 
struts  i4,  M,  Z6,  m8,  etc. 

=  8  +  *g  +  30  sec  0  =  25  x  1.414  =  35.4  tons. 

The  maximum  stress  in  the  lower  part  of  any  main  tie, 
as  1 8,  occurs  when  the  excess  loads  act  at  8  and  at  e,  48  feet 
to  the  right  of  8,  and  the  train  loads  are  at  all  the  apexes 


150  ROOFS  AND  BRIDGES. 

from  the  right  abutment  to  8  inclusive.     The  shear  in  18  for 
this  loading  is, 

2*  x  8  +  f  £  (io  +  7)  +||  (10  +  -  +  1)  =  93.125. 
.-.  max.  stress  in  IS  =  93.125  x  1.414  =  131.7  tons. 

The  maximum  stress  in  the  upper  part  of  any  main  tie, 
as  3  A:,  occurs  when  the  excess  loads  act  at  6  and  at  d,  48 
feet  to  the  right  of  6,  and  the  train  loads  are  at  all  the 
apexes  from  the  right  abutment  to  b  inclusive  :  for  half  of 
the  train  load  at  b  is  transmitted  to  the  joint  6  through  the 
half  strut  k  6,  and  -^  of  the  train  load  at  b  is  transmitted  to 
the  right  abutment  ;  therefore  the  difference  of  these,  or  -fir 
of  the  train  load  at  b,  goes  to  the  left  abutment  and  forms 
part  of  the  shear  in  3fc;  that  is,  the  train  panel  load  at  6 
will  increase  the  shear  in  3k  by  ^  x  12  =  3f  tons.  It  is 
evident  also  that  half  of  the  dead  load  at  6,  which  is  trans- 
mitted to  6  through  the  half  strut  k  6,  goes  to  the  left  abut- 
ment and  forms  part  of  the  shear  in  3  k.  Similarly  for  5  1 
and  7m.  Hence, 

Max.  stress  in  3  A; 

=  [5x8  +  f£(12  +  9)  +  if  (12  +  .»  +  1)  +  3.75]  1.414 
=  200.3  tons. 

The  maximum  stress  in  3-4  is  equal  to  the  full  panel  load 
at  4  plus  half  the  sum  of  the  panel  loads  at  a  and  b 
=  30  +  12  +8  +|(24  +  16)  =  70  tons. 

The  maximum  stress  in  either  of  the  other  long  verticals, 
as  7-8,  is  equal  to  the  shear  in  7  m  plus  whatever  load  is 
applied  to  the  upper  apex  7.  Hence, 

Max.  stress  7-8 


=  -60.1  tons. 


BRIDGE   TKUSSES. 


151 


The  maximum  stress  in  3 1  occurs  when  the  excess  loads 
act  at  4  and  at  c,  and  the  train  loads  cover  the  whole  truss ; 
it  is  easily  seen  that  half  of  the  dead  train  loads  at  a, 
which  are  transmitted  to  4,  form  part  of  the  shear  in  3  i. 
Hence, 

Max.  stress  in  3  i  =  -  [7  x  20  +  f  £  (14  +  11)]  1.414 

=  -  264.2  tons. 

The  following  stresses  are  found  in  a  manner  similar  to 
the  above : 

MAXIMUM  CHORD  STRESSES. 


3-5 

5-7 

7-9 

2-4 

4-6 

6-8 

8-10 

-318.8 

-395.6 

-417.5 

+  196.9 

+  196.9 

+328.8 

+405.6 

MAXIMUM  STRESSES  IN  LOWER  PART  OF  MAIN  AND  HALF  DIAGONALS. 


«-2 

£-6 

1-8 

»n-10 

i-4 

k^ 

/-6 

m-8 

-283.7 

+189.3 

+  131.7 

+  78.3 

-35.4 

-35.4 

-35.4 

-35.4 

MAXIMUM  STRESSES  IN  UPPER  PART  OF  MAIN  DIAGONALS. 


3-i 

3-/fc 

5-1 

1-m 

9-TO 

1-1 

-264.2 

+200.3 

+  140.5 

+  85.0 

+53.2 

+6.2 

MAXIMUM  STRESSES  IN  THE  VERTICALS. 


3-4 

6-6 

7-8 

9-10 

ai,  bk,  cl,  dm. 

+  70.0 

-99.4 

-60.1 

-23.9 

+  50.0 

152  ROOFS  AND  BRIDGES. 

The  stress  in  9-m  =  9-n  occurs  when  the  excess  loads 
act  at  e  and  6',  and  the  train  loads  act  at  all  the  joints 
from  the  right  abutment  to  e.  Then  the  shear  in  the  sec- 
tion cutting  9-7',  9-n,  and  10-n  =  101.625  —  8x8=+  37.625 
tons;  and  as  10-n  cannot  take  compression,  this  is  the 
vertical  component  of  stress  in  9-n. 

.-.  max.  stress  in  9-n  =  +  37.625  x  1.414  =  +  53.2  tons. 

If  10-n  were  constructed  to  take  compression,  this  would 
he  greatly  modified.  In  this  case  the  excess  loads  would 
act  at  8'  and  g,  and  the  train  loads  at  all  the  joints  from  the 
right  abutment  to  8'.  Then  the  left  reaction  =  92.625  tons 
and  the  shear  in  the  section  cutting  9-7',  9-n,  and  10-n  = 
92.625  -  64  =  +  28.625  tons.  The  member  10-n  takes  a 
positive  shear  of  4  tons,  so  that  the  vertical  component  of 
stress  in  9-n  =  +  28.625  -  4  =  +  24.625  tons. 

.-.  max.  stress  in  9-n  =  +  24.625  x  1.414  =  +  34.8  tons. 

The  stress  in  7-0  occurs  when  the  excess  loads  act  at 
/  and  4',  and  the  train  loads  at  all  the  joints  from  the 
right  abutment  to  /.  Then  the  left  reaction  =  +  84.375  tons 
and  the  shear  in  the  section  cutting  7 '-5',  7'-0,  and  8'-0  = 
84.375-  8  x  10  =  +4.375  tons.  As  8'-0  does  not  act  in 
compression,  this  shear  must  be  resisted  by  7'-0  alone. 

.-.  max.  stress  in  7'-0  =  +  4.375  x  1.414  =  +  6.2  tons.* 

Prob.  117.  A  deck  Baltimore  truss,  Fig.  54,  has  16  panels, 
each  10  feet  long  and  20  feet  deep,  the  verticals  being  posts 
and  the  diagonals  ties ;  the  dead  and  train  loads  are  1000 
Ibs.  and  2000  Ibs.  per  foot  per  truss,  and  there  are  two 
excess  loads  50  feet  apart,  each  of  33  tons :  find  the  stresses 
in  all  the  members. 

*  It  should  be  noted  that  the  members  m-S  and  n-8'  are  also  tension 
members  for  counter  stresses. 

The  tension  stress  in  each  of  these  members  =  20.625  X  1.414  =  29.2  tons. 


BRIDGE  TRUSSES. 
MAXIMUM  CHORD  STRESSES. 


153 


1-3 

3-5 

5-7 

7-9 

2-1 

4-6 

6-3 

-164.1 

-274.1 

-337.7 

-354.7 

+  152.4 

+258.4 

+317.8 

MAXIMUM  STRESSES  IN  UPPER  PART  OF  MAIN  AND  HALF  TIES. 


H 

s-i 

5-/fc 

l-l 

3-t 

5-J 

1-k 

9-Z 

+232.0 

+  180.6 

+  132.7 

+88.3 

+34.0 

+34.0 

+34.0 

+47.4 

MAXIMUM  STRESSES  IN  LOWER  PART  OF  MAIN  TIES. 


i-2 

.M 

£-6 

M 

6-1 

+  215.5 

+  164.1 

+  116.2 

+69.2 

+  31.8 

The  counter  4-fc  is  found  not  to  be  required ;  but  in  actual 
practice  a  counter  would  be  used  to  provide  for  loads  differ- 
ing from  the  ones  above  assumed. 

MAXIMUM  STRESSES  IN  THE  VERTICALS. 


2-3 

4-5 

6-7 

8-9 

ni,  hi,  ck,  dl. 

-152.4 

-116.1 

-82.2 

-69.2 

-48.0 

This  problem  is  taken  from  "Strains  in  Framed  Structures"  by 
A.  J.  Du  Bois.  It  will  be  seen  that  the  answers  above  given  for  the 
maximum  stresses  in  the  lower  part  of  the  main  ties  and  in  the  Ion  ; 
verticals  differ  somewhat  from  those  given  by  Prof.  Du  Bois. 


154  ROOFS  AND  BRIDGES. 

Art.  44.  True  Maximum  Shears  for  Uniform  Live 
Load.  —  Thus  far,  it  has  been  assumed  that  the  live  panel 
loads  were  all  concentrated  at  the  panel  points,  and  in  order 
to  find  the  maximum  stress  in  any  diagonal  due  to  a  uniform 
live  load  per  linear  foot,  the  maximum  live  load  shear  in 
that  diagonal  has  been  found  by  putting  a  live  panel  load 
on  every  joint  on  the  right  of  the  section  cutting  the  diagonal 
(Art.  22). 

For  example,  it  has  been  assumed  that  the  first  apex  on 
the  right  of  the  section  in  Fig.  57  has  a  full  live  panel  load 
and  the  first  one  on  the  left  has  no  live  load.  Now  the  joint 
n  on  the  right  of  the  section  receives  the  half  panel  load 

AAA 


A 
T 


ai 

-» 


R  -  -----     ---- 

Fig.  GT 

between  the  joints  n  and  n  —  1,  but  only  a  small  part  of  the 
load  on  the  left  of  n  ;  the  other  part  of  the  load  on  the  left 
of  n  is  received  by  the  joint  n  +  1.  When  the  uniform  load 
extends  from  the  right  abutment  to  the  joint  n  -f  1,  the  joint 
n  receives  a  full  panel  load  ;  but  then  the  joint  n-f-1  receives 
a  half  panel  load.  Let  us  therefore  ascertain  at  what  dis- 
tance x  to  the  left  of  the  joint  n  the  load  must  extend  to 
produce  the  greatest  shear  in  the  n  -f  1th  panel. 

Let  N—  the  number  of  panels  in  the  truss,  p  =  the  panel 
length,  w  =  the  uniform  live  load  per  linear  foot,  n  =  the 
number  of  whole  panels  covered  by  the  load,  R  =  the  reac- 
tion at  the  left  or  unloaded  end.  Then, 

^-  =  the  part  of  the  load  wx  which  is  carried  by  the  joint 

2p  <> 

n  +  1  ;  and  wx  —  ^^-  =  the  part  of  the  load  wx  which  is 

2p 
carried  by  the  joint  n. 


BRIDGE   TRUSSES.  155 

We  have  then  for  the  reaction 


wxn       wx2        ujn  p  ,-,  x 

N  ^2N  +  2N' 


Hence  the  shear  in  the  n  +  1th  panel  is 

-p.  _  ztmi        wee2       wnzp     wx2  ,n\ 

*  XT      2pN     2ff     2p' 


Equating  the  derivative  of  V  to  zero,  we  have, 


which  is  the  distance  to  the  left  of  the  joint  n  that  the  live 
load  must  extend  in  order  to  produce  the  maximum  shear 
in  the  n  +  1th  panel  from  the  right  end. 

Thus,  let  the  truss  have  8  panels ;  then  for  the  true 
maximum  shear  in  the  sixth  panel  from  the  right  end 
x  =  $p,  for  the  seventh  panel  x  =  $p,  and  for  the  eighth  or 
last  panel  x  =  ip  —p,  or  the  whole  truss  is  covered. 

The  total  live  load  on  the  truss 


=  N  times  the  load  on  the  n  +  1th  panel. 

Therefore,  the  live  load  on  the  n  +  1th  panel  is  —  th  of  the 
total  live  load  on  the  truss. 

Substituting  the  above  value  of  x  in  (2)  and  reducing,  we 

have  Maximum  shear  =  ,  ,«*"»'     . 


156  ROOFS  AND  BRIDGES. 

These  values  of  x  and  the  maximum  shear  are  independent 
of  the  form  of  the  truss,  whether  the  webbing  be  inclined, 
as  in  the  Warren,  or  inclined  and  vertical,  as  in  the  Pratt 
and  Howe. 

Prob.  118.  A  truss,  Fig.  57,  has  7  panels,  each  16  feet 
long,  the  live  load  is  2  tons  per  foot:  find  (1)  the  true 
maximum  live  load  shears  in  every  panel  and  (2)  the 
maximum  live  load  shears  by  the  usual  method  (Art.  22). 

Here  N=  7,  p  =  16  feet,  and  w  =  2  tons. 

For  the  maximum  shear  in  the  1st  panel  from  the  left 
w  =  6. 

.-.  x  =  p  =  16  feet,  and  shear  =  2  x  16  x  ®  =96  tons, 

^    X    O 

which  is  half  of  the  effective  load,  or  the  effective  reaction. 

For  max.  shear  in  the  4th  panel  from  left,  n  =  3. 
.«,  x  =  f  x  16  =  8  feet,  or  the  load  reaches  to  the  middle 
of  the  4th  panel. 

Shear  =  2  X  16  X  <S*  =  24.0  tons. 
2x6 

By  the  usual  way  of  taking  a  full  panel  load  as  con- 
centrated at  the  third  apex  from  the  right  n  —  1,  and  none 
at  the  4th  apex  n,  we  have 

Shear  =  -^  (1  +  2  +  3)  =  27.4  tons, 

which  is  greater  than  that  obtained  by  the  strictly  correct 
method. 

Thus  the  following  live  load  shears  are  computed  (1)  by 
the  Strictly  Correct  Method  and  (2)  by  the  Method  in  Common 
Practice  : 

True  Method.  96.0,  66.7,  42.7,  24.0,  10.7,  2.7,  0.0  tons. 
Common  Method.  96.0,  68.6,  45.7,  27.4,  13.7,  4.6,  0.0  tons. 


BEIDGE   TRUSSES.  157 

It  is  seen  that  the  shears  obtained  by  the  usual  method, 
that  is,  by  supposing  the  panel  loads  to  be  concentrated  at 
the  panel  points,  are  larger  than  those  obtained  by  the  true 
method.  For  the  reason  that  the  shears  obtained  by  the 
usual  method  always  err  on  the  safe  side,  and  that  the  error 
is  always  small,  it  is  the  common  practice  to  suppose  all  the 
load  from  middle  to  middle  of  panel  to  be  concentrated  at 
the  panel  point,  as  in  Art.  4. 

Art.  45.  Locomotive  Wheel  Loads.  —  In  computing 
the  stresses  in  railway  bridges  in  America,  the  live  load 
which  is  now  generally  taken,  consists  of  two  of  the  heaviest 
engines  in  use  on  the  line,  at  the  head  of  the  heaviest 
known  train  load.  The  weights  of  the  engines  and  tenders 
are  assumed  to  be  concentrated  at  the  wheel-bearings, 
giving  definite  loads  at  these  points,  while  the  train  load 
is  taken  as  a  uniform  load  of  about  3000  pounds  per  linear 
foot  of  single  track. 

Oj  ^  Oj  C\j 

oo    OO    oooo    oo    OO    0000 
Ui-9'4"*' 


(a.) 

o  o     o  §      <o       JSojjucStvj      §  §    §  g 
n  o    n  o      n      OOQOO     o  n    n  o 


(  b  )  (UOffet  —  1  inch  scale) 

Fig.  58. 

The  first  diagram  of  Fig.  58  represents  two  88-ton  pas- 
senger locomotives,  as  specified  by  the  Pennsylvania  Rail- 
road. 

The  second  diagram  shows  two  112-ton  decapod  engines, 
used  on  the  Atlantic  Coast  Line. 


158 


ROOFS  AND   BRIDGES. 


The  numbers  above  the  wheels  show  their  weights  in  tons 
for  both  rails  of  a  single  track.  The  numbers  between  the 
wheels  show  their  distances  apart  in  feet. 

Art.  46.  Position  of  Wheel  Loads  for  Maximum 
Shear.  —  Let  Fig.  59  represent  a  truss  with  a  decapod 
engine  and  train  upon  it. 

Let  N  =  the  number  of  panels  in  the  truss,  p  =  the  panel 
length,  P=  Pl  +  P2  +  •••  +P10  =  the  sum  of  the  wheel 
loads,  w  =  the  uniform  train  load  per  linear  foot,  W=  the 
total  live  load  on  the  truss,  d  =  the  distance  of  the  center  of 


gravity  of  P  from  the  front  of  the  train,  x  =  the  distance  of 
the  front  of  the  train  from  the  right  abutment,  a  =  the  dis- 
tance of  Pl  from  the  front  of  the  train,  y  =  its  distance  at 
the  left  of  the  nth  panel  point,  and  R  =  the  reaction  at  the 

left  end.     Then  Pl  #  =  the  part  of  the  load  P^  that  is  carried 

P 
by  the  n  -f  1th  panel  point. 

We  have  then  for  the  reaction 


i. 

Np       '  2Np 
Hence  the  shear  in  the  n  +  1th  panel  is 


(1) 


BRIDGE  TRUSSES.  159 

But  from  the  figure,  x  +  a  =  np  +  y  ;  .'.  y  =  a  +  x  —  np, 
which  in  (2)  gives 


Equating  the  first  derivative  of  F  to  zero,  we  have, 
P  4-  wx      p  _  n 

~^r~  yi~°' 

or,  since  P  +  wx=W,  we  have, 


Hence,  the  shear  in  any  panel  is  a  maximum  when  the  load 
on  the  panel  is  —  th  of  the  entire  live  load  on  the  truss. 

In  practice  it  is  convenient  to  put  one  of  the  loads  at  the 
nth  panel  point,  so  that  the  above  condition  cannot,  in  gen- 
eral, be  exactly  satisfied.  We  must  have  in  general, 


Hence,  in  general,  the  shear  in  the  n  +  1th  panel  ts  a 
maximum  when  one  of  the  loads  is  at  the  nth  panel  point,  and 
the  load  on  the  n  +  It  h  panel  is  equal  to  or  just  less  than  —  th 
of  the  entire  load  on  the  truss.1 

Prob.  119.  A  through  Warren  truss  has  10  panels,  each 
12  feet  long  :  find  the  maximum  shears  in  each  panel  caused 
by  a  single  decapod  engine  and  tender. 

Here  each  driver  in  Fig.  59  weighs  12.8  tons,  each  tender 
wheel  weighs  10  tons,  and  the  pilot  wheel  weighs  8  tons; 
the  total  load  W  is  112  tons. 

1  This  holds  good  in  all  cases  for  shear. 


160  ROOFS  AND  BRIDGES. 

Let  it  be  required  to  find  the  maximum  shear  in  the  2d 
panel  from  the  left.  Here  n  =  8  ;  N=W. 

By  the  above  rule  for  the  maximum  shear  in  this  panel, 
the  first  driver  must  be  put  at  the  joint  n,  since  8  <  Ly,  and 


. 
In  this  position  of  the  loads,  the  left  reaction  is 

o  1  9  Q 

fi=  -5-  x  104+  ^  (96.0  +  91.75  +  87.5  +  83.25  +  79.0) 

+  _12_  (71.5  +66.83  +61.23  +56.56)  =74.9  tons. 
.-.  max.  shear  =  74.9  -  ^-~  =  69.6  tons. 

Let  it  be  required  to  find  the  maximum  shear  in  the  8th 
panel  from  the  left.     Here  n  =  2. 

To  find  the  position  of  the  wheels  for  maximum  shear  in 
this  panel,  try  the  first  driver  at  the  joint  n,  as  before. 

For  this  position  the  total  live  load  on  the  truss 

=  8  +  5  x  12.8  =  72  tons, 

since  the  tender  wheels  are  off  the  truss.     By  the  rule, 
therefore,  this  is  not  the  correct  position,  since  8  >  -^  x  72. 

If  we  put  the  pilot  wheel  at  n,  the  load  on  the  truss 

=  8  +  4  x  12.8  =  59.2  tons, 

since  the  last  driver  is  off  the  truss.     Hence  for  maximum 
shear  in  this  panel  this  is  the  correct  loading. 

With  the  live  load  in  this  position,  the  left  reaction  is 

R  =  ifo  x  24  +  m  (16  +  1L75  +  7'50  +  3'25)=  5'7  tons' 

which  is  also  the  maximum  shear,  since  in  this  case  there  is 

no  load  to  be  subtracted. 

Thus  the  following  shears  are  determined  : 

Max.  shears  =  80.8,  69.6,  58.4,  47.2,  36.0,  24.8,  13.9,  5.7. 


BRIDGE   TRUSSES.  161 

Prob.  120.  A  deck  Pratt  truss,  Fig.  50,  has  10  panels, 
each  20  feet  long  and  24  feet  deep  :  find  the  maximum  web 
stresses  in  each  panel  caused  by  a  single  passenger  locomo- 
tive and  tender,  as  shown  in  (a)  of  Fig.  58. 

We  first  find  the  maximum  shear  in  the  left  panel.  l>_y 
the  rule  the  second  pilot  wheel  must  be  put  at  the  joint  3. 
since  8  <  f  f  ,  and  8  +  8  >  f  f  . 

With  the  live  load  in  this  position,  the  left  reaction  is 

R  =  A  (185.5  4-  180  +  153.5  +  148.5  +  143  +  138) 
200 


+        (171  +  163)  =71.34  tons; 
and  the  max.  shear  =  71.34  -  8  x  |^  =  69.14. 

.-.  stress  in  1-4  =  69.14  sec  6  =  69.14  x  1.302  =  90.0  tons. 

For  3-4  we  have  the  same  loading  as  for  1-4  ;  and  the 
maximum  compression  in  3-4  is  equal  to  the  shear  just 
found  for  1-4. 

.-.  stress  in  3-4  =  —  69.1  tons. 

The  maximum  compression  for  the  end  vertical  1-2  is 
found  by  placing  the  wheels  so  as  to  bring  the  greatest  load 
to  the  joint  1,  and  is  found  to  be  71.9  tons. 

Thus  we  find  the  following  maximum  stresses  : 

Diagonals  =  +  90.0,  +78.6,  +67.1,  +55.6,  +44.2,  +32.7, 

+  21.2,  +  9.8  tons. 
Posts  =  -77.7,  -69.1,  -60.3,  -51.5,  -42.7,  -33.9  tons. 

If  the  second  driver  be  placed  at  the  middle  panel  point  11,  the 
stress  in  11-12  will  be  found  to  be  -  39.6  tons. 

Prob.  121.  A  deck  Pratt  truss,  Fig.  50,  has  10  panels, 
each  12  feet  long  and  12  feet  deep  ;  the  dead  load  is  1000 


162 


ROOFS  AND   BRIDGES. 


Ibs.  per  linear  foot  of  track,  the  live  load  is  a  passenger 
locomotive  and  tender,  as  shown  in  (a)  of  Fig.  58 :  find  the 
maximum  web  stresses  in  each  panel. 

The  dead  and  live  load  stresses  may  be  computed  sepa- 
rately. 

STBESSES  IN  DIAGONALS. 


MEMBERS. 

1-4 

3-6 

5-8 

7-10 

9-12 

11-10' 

9'-3' 

Dead  stresses. 
Live  stresses  . 

+  38.0 

+  88.3 

+  29.6 
+  75.9 

+21.1 
+63.4 

+  12.7 
+  51.0 

+  4.2 
+  38.6 

-  4.2 
+26.1 

-12.7 
+14.8 

Max.  stresses. 

+  126.3 

+  105.5 

+  84.5 

+63.7 

+42.7 

+21.9 

+  2.1 

STRESSES  IN  THE  POSTS. 


MEMBERS. 

1-2 

3-4 

e-6 

7-3 

9-10 

•11-12 

Dead  stresses  . 
Live  stresses  . 

-  30.0 
-  70-9 

-27.0 
-62.5 

-21.0 
-53.7 

-15.0 
-44.9 

-  9.0 
-36.1 

-  6.0 
-27.3 

Max.  stresses  . 

-100.9 

-89.5 

-74.7 

-59.9 

-45.1 

-33.3 

If  the  first  driver  be  placed  at  the  middle  panel  point  11,  the  stress 
in  11-12  will  be  found  to  be  -  28.7  tons.  It  will  be  noticed  that  the 
tension  of  4.2  tons  in  the  diagonal  9'-12  is  equivalent  to  a  compression 
of  4.2  tons  in  the  diagonal  11-10'  ;  and  similarly  for  the  next  diagonal 


Prob.  122.    A  through  Pratt  truss,  Fig.  60,  has  8  panels, 
each  18  feet  long  and  24  feet  deep  :  find  the  maximum  web 


5" 3' 


BRIDGE  TRUSSES.  163 

stresses  in  each  panel  due  to  a  passenger  locomotive  and 
tender,  as  shown  in  (a)  of  Fig.  58,  followed  by  a.  uniform 
train  load  of  3000  Ibs.  per  linear  foot. 

(1)  To  find  the  maximum  stress  in  2-3. 

Try  the  first  driver  at  the  panel  point  4.  In  this  position 
the  total  live  load  on  the  truss 

=  88  4-  90  x  1.5  =  223  tons. 

Since    8+8<Af*,   and   8+8  +  20>^p,  this    is  the 
correct  position  for  the  maximum  shear  in  the  panel  2-4. 
The  reaction  is 

R  =  —  (135  +  140.5  + 108.5  +  103.5  +  98  +  93) 

144 

+  ^(126  +  118)+  \  x  1^°  =  113.77  tons. 

.-.  max.  shear  =  113.77  -  —  (9  +  14.5)=  103.33. 
18 

.-.  stress  in  2-3  =  -  103.33  x  1.25  =  -  129.2  tons. 

(2)  To  find  the  maximum  stress  in  9-8'. 

Try  the  first  driver  at  the  joint  8'.  In  this  position  the 
total  live  load  on  the  truss 

=  88  +  18  x  1.5  =  115  tons. 

But  8  +  8  >  -i-^ ;  hence  this  is  not  the  correct  position 
for  the  maximum  shear  in  the  panel  10-8'. 

We  therefore  try  the  second  pilot  wheel  at  8'.  We  then 
have,  for  the  total  live  load  on  the  truss, 

W=  88  +  9  x  1.5  =  101.5  tons. 

Since  8  <  — -^-,  this  is  the  correct  position  for  the  maxi- 

8 
mum  shear  in  the  panel  10-8'. 


164  ROOFS  AND  BRIDGES. 

The  reaction  is 

R  =  1540  +  1640  +  13.5x4.5 
144 

.-.  max.  shear  =  22.51  -  ^^  =  20.1  tons. 

lo 
=  —  stress  in  9-10. 

.-.  stress  in  9-8'  =  20.1  x  1.25  =  25.1  tons. 

The  max.  tension  in  the  hip  vertical  3-4  is  found  by 
putting  the  wheels  so  as  to  bring  the  greatest  load  at  4. 
Thus  the  following  maximum  stresses  are  found  : 
Diagonals 

=  -  129.2,  +  96.4,  +  67.9,  +  43.6,  +  25.1,  +11.3  tons. 
Verticals  =  +  36.9,  -  54.3,  -  34.9,  -  20.1  tons. 

Art.  47.  Position  of  Wheel  Loads  for  Maximum 
Moment  at  Joint  in  Loaded  Chord.  —  In  addition  to 
the  notation  of  Art.  46,  let  P  =  the  load  on  the  left  of  the 
panel  point  n,  Fig.  59,  <c'  =  the  distance  of  its  center  of 
gravity  from  the  point  n,  and  n'  —  the  number  of  panels 
between  the  left  abutment  and  the  point  n.  Then  for  the 
moment  at  the  panel  point  n,  Fig.  59,  we  have 

M=Rn'p-Px' 


Of  Art  461 
Equating  the  first  derivative  of  M  to  zero,  we  have,  since 


or  since  P+  wx  =  W  (Art.  46),  we  have 

p.-'w. 


BEIDGE  TRUSSES.  165 

Hence  the  moment  at  any  panel  point  in  the  loaded  chord  is 

n' 
a  maximum  when  the  load  on  the  left  of  the  point  is  — ths  of 

the  entire  live  load  on  the  truss. 

In  practice  it  is  convenient  to  put  one  of  the  loads  at  the 
nth  panel  point,  so  that  the  above  condition  can  very  seldom 
be  exactly  satisfied.  We  must  have  in  general 

P'  =  or<  —  W. 

N 

Hence,  in  general,  the  moment,  at  any  panel  point  of  the 
loaded  chord  is  a  maximum  when  the  load  on  the  left  of  the 
point  has  to  the  entire  load  on  the  truss  a  ratio  ivhich  is  equal 
to  or  just  less  than  the  ratio  which  the  number  of  panels  on 
the  left  of  the  point  bears  to  the  entire  number  of  panels  in  the 
truss. 

By  this  rule  the  maximum  chord  stress  in  any  member  of 
the  unloaded  chord  may  be  determined,  since  we  have  only 
to  divide  the  maximum  moment  at  the  panel  point  opposite 
the  chord  member  by  the  depth  of  truss. 

Prob.  123.  A  deck  Pratt  truss,  like  Fig.  50,  has  8  panels, 
each  18  feet  long  and  24  feet  deep:  find  the  maximum 
chord  stresses  due  to  a  single  passenger  locomotive  and 
tender. 

Let  it  be  required  to  find  the  maximum  stress  in  the 
second  panel  3-5.     Here  n'  =  2,  N=  8,  W=  88 ;  therefore, 
by  the  rule,  P  must  be  equal  to  or  less  than  f  x  88,  or  22 
tons.     Hence,  the  first  driver  must  be  put  at  5  since 
8+8<22,  and8  +  8  +  20>22. 

With  the  live  load  in  this  position,  the  left  reaction  is 
E  =  —  (122.5  + 117  +  90.5  +  85.5  +  80  +  75) 

90 

100)  =  60.58  tons. 


166  ROOFS  AXD  BRIDGES. 


.-.stress  in  3-5  =  ---  -       =  -83  tons 

=  —  stress  in  6-8. 

Thus  are  found  the  following  stresses  : 
Max.  stress  in  1-3=  —47.7,  in  3-5=  —83,  in  5-7=  —103.8, 
in  7-9  =  —  110.5  tons. 

Prob.  124.  A  deck  Pratt  truss,  Fig.  50,  has  10  panels, 
each  20  feet  long  and  24  feet  deep:  find  the  maximum 
chord  stresses  in  each  panel  due  to  a  passenger  locomotive 
and  tender. 

Ans.   Max.  stresses  in  upper  chord  =  —  57.6,    —  103.0, 

-  136.4,  -  155.2,  -  161.9  tons. 

Prob.  125.  A  through  Warren  truss  has  10  panels,  each 
12  feet  long  and  12  feet  deep  :  find  the  maximum  stresses 
in  the  upper  chord  due  to  a  decapod  engine  and  tender. 

Ans.  Max.  stresses  =  -  80.8,  —145.1,  -190.4,  -217.3, 

-  225.2  tons. 

Art.  48.  Position  of  Wheel  Loads  for  Maximum 
Moment  at  Joint  in  Unloaded  Chord.  —  By  the  rule 
deduced  in  Art.  47,  the  maximum  chord  stresses  may  be 
determined  in  the  unloaded  chord  of  any  simple  truss,  and 
also  in  the  loaded  chord  of  such  trusses  as  the  Pratt  and 
Howe,  where  the  web  members  are  vertical  and  inclined  so 
that  the  panel  points  of  the  upper  chord  are  directly  over 
those  of  the  lower  chord.  For  trusses  like  the  Warren  and 
lattice,  where  all  the  web  members  are  inclined,  it  applies 
only  to  the  unloaded  chord.  For  the  loaded  chord  of  such 
trusses  a  modification  of  the  formula  or  rule  is  necessary, 
which  may  be  deduced  as  follows  : 

Let  it  be  required  to  find  the  maximum  moment  at  the 
panel  point  c  of  the  unloaded  chord,  Fig.  59. 


BRIDGE   TRUSSES.  167 

Let  W=  the  total  live  load  on  the  truss,  Q  =  the  load 
on  the  (n  —  l)th  panel,  and  P  =  the  load  on  the  left  of 
the  (n  —  l)th  panel. 

Let  V  =  the  distance  of  c  from  the  left  support,  q  =  the 
distance  be,  p  =  the  panel  length,  and  I  =  the  length  of  the 
span. 

Let  x  =  the  distance  from  the  center  of  gravity  of  W  to 
the  right  support,  ccj=  the  distance  from,  the  center  of  gravity 
of  P'  to  the  panel  point  n,  and  x2  =  the  distance  from  Q  to 
the  panel  point  n  —  1. 

The  part  of  Q  that  is  carried  by  the  nth  panel  point 


Hence  the  moment  at  c  is 


Equating  the  first  derivative  of  M  to  zero,  we  have,  since 
dx  =  dxl  =  dx2, 

J^_P'_Q2  =  o.    .'.P>  +  ZQ  =  ±W.    .   .   .   (i) 

If  the  braces  are  equally  inclined,  that  is,  if  the  center 
of  moments  c  is  directly  over  or  under  the  center  of  the 
opposite  member,  as  is  usually  the  case,  we  have  -  =  -,  and 
(1)  becomes  ^ 

W:       ......      (2) 


For  the  Pratt  and  Howe  trusses  q  =  0,  and  (1)  becomes 


which  is  the  same  as  the  formula  that  was  found  in  Art.  47. 


168  ROOFS  AND  BRIDGES. 

It  is  convenient  in  practice  to  put  one  of  the  loads  at  the 
n  —  1th  panel  point,  so  that,  in  general,  we  must  have 


(3) 


Prob.  126.  A  through  Warren  truss  has  10  panels,  each 
12  feet  long  and  12  feet  deep:  find  the  maximum  stresses 
in  the  lower  chord  due  to  a  decapod  engine  and  tender. 

To  find  the  maximum  stress  in  the  first  panel  2-4.  Here 
Z'=L,  Z  =  10,  and  TF=112.  Therefore  by  formula  (3), 
P  +  £  Q  must  be  equal  to  or  less  than  ^  x  112  or  5.6  tons. 
Hence  the  1st  driver  must  be  put  at  4,  since 

|  X  8  <  5.6,  and  8  +  1  x  12.8  >  5.6. 

The  left  reaction  then  =  86.14  tons. 
The  moment  at  the  point  1  is 

M  =  86.14  x6-8x^-x6  =  484.84. 


.-.  stress  in  2-4  =  =  40.4  tons. 

Similarly  the  maximum  stresses  in  the  other  lower  chord 
members  are  found. 

Max.  stresses  =  40.4,  111.9,  165.4,  201.7,  219.0  tons. 

Sug.  To  find  stress  in  4-6,  put  2d  driver  at  6  ;  in  6-8,  put  3d 
driver  at  8  ;  in  8-10,  put  4th  driver  at  10  ;  in  10-12,  put  5th  driver 
at  12. 

Art.  49.  Tabulation  of  Moments  of  Wheel  Loads. 

—  A  diagram  such  as  is  shown  in  Fig.  61,  diminishes  con- 
siderably the  work  of  computing  stresses  due  to  actual  wheel 
loads.  The  first  diagram  is  for  an  88-ton  passenger  loco- 
motive and  its  tender  ;  the  second  is  for  a  112-ton  decapod 
engine  and  its  tender.  Any  locomotive  can  be  diagramed 


BRIDGE   TRUSSES. 


169 


in  the  same  manner ;  and  the  same  method  applies  to  two 
locomotives  coupled  together. 


Q 0 


a 
O 


o   o    o    c> 


1  s.'5  la- 

P-*K  - 


*        §E          |J   |!| 

K----56  75/75,.  x.'f 4  5;£    ^jfc 

U - 64Tons;  32'.0 * 

N 72T&/75,  37fr  •— >j 

K 88Tons,47!s 

H W7&/75;  5H'5 


(a.) 

I  I  1 1  I 

00000    oono 


K SZZToris, 

'f-* 72Tonsi  25.'0  ••; — •>! 

S 102  fans,  42^77 ->i        !    1 

K lIZTbmj  47.43;~*~ -' >«    J 

(b.) 
Fig.  61. 
(20  feet  =  1  inch  scale.) 

Each  diagram  shows  the  weights  and  distances  apart  of 
the  wheels.  Below  each  wheel,  on  the  horizontal  line,  is 
shown  the  weight  of  that  wheel  together  with  that  of  all 
the  preceding  ones,  and  its  distance  from  the  front  wheel. 


170  ROOFS  AND  BRIDGES. 

Below  each  wheel  on  the  vertical  line  is  given  the  moment 
of  all  the  preceding  wheels,  with  reference  to  that  wheel. 
Thus,  for  the  third  driver  of  the  decapod  engine,  we  have 
46.4  tons  for  the  weight  of  it  and  the  three  preceding 
wheels,  16.5  feet  for  its  distance  from  the  front  wheel,  and 
295.2  foot-tons  for  the  moment  of  the  preceding  wheels  with 
reference  to  the  third  driver.  At  the  beginning  of  the 
uniform  load  we  have  112  tons  for  the  weight  of  all  the 
preceding  loads,  49.68  feet  for  the  distance  of  this  point 
from  the  front  wheel,  and  2910  foot-tons  for  the  moment  of 
all  the  preceding  loads  with  reference  to  this  point.  Each 
diagram  shows  also  that,  the  moment  at  any  tvheel  is  equal  to 
the  moment  at  the  next  preceding  wheel  on  the  left,  plus  the 
sum  of  all  the  preceding  ivheel  loads  multiplied  by  the  distance 
from  the  next  left  preceding  wheel  to  the  wheel  in  question. 
Thus,  in  (a)  the  moment  of  the  first  three  wheels  about  the 
third  is  188,  and  about  the  fourth  it  is  476.  But  476  is  equal 
to  188  plus  36  multiplied  by  8;  and  similarly  for  the 
moment  with  reference  to  any  other  point.  (See  Du  Bois's 
Strains  in  Framed  Structures.) 

This  diagram  may  be  used  for  finding  reactions,  shears, 
and  moments.  Thus 

Let  I  =  the  length  of  the  truss,  and  the  other  notation  as 
in  Art.  46.  Then,  from  (1)  of  Art.  46,  the  left  reaction  is 


I 

which  for  the  passenger  locomotive  is 

and  for  the  decapod  engine  is 

r29iO  +  U2*+5«"*V    •    •    •     (2) 


HO 


BRIDGE  TRUSSES.  171 

Suppose,  for  example,  that  the  third  tender  wheel  of  the 
decapod  engine  is  2  feet  to  the  left  of  the  right  abutment. 
We  take  out  the  numbers  2181.8  and  102  from  the  diagram, 
and  the  reaction  is 

S =i  (2181.8  + 102  x  2). 

In  practice  it  is  convenient  to  draw  a  skeleton  outline  of 
the  truss  to  the  same  scale  as  the  diagram,  to  be  placed 
directly  above  it  in  the  proper  position  for  the  maximum 
stress  in  each  member. 

Prob.  127.  If  the  span  be  120  feet,  find  the  reactions  for 
the  passenger  locomotive,  (1)  when  the  last  tender  wheel  is 
5  feet  to  the  left  of  the  right  abutment,  (2)  when  the  first 
driver  is  50  feet  to  the  right  of  the  left  abutment,  and 
(3)  when  the  second  pilot  wheel  is  40  feet  to  the  left  of  the 
right  abutment. 

Ans.    (1)  21.4  tons ;    (2)  52.1  tons ;    (3)  16.4  tons. 

Prob.  128.  If  the  span  be  100  feet,  find  the  moments  for 
the  decapod  engine,  (1)  at  an  apex  30  feet  to  the  left  of  the 
right  abutment  when  the  second  driver  is  at  that  apex,  and 
(2)  at  the  center  of  the  span  when  the  first  driver  is  there. 

Ans.  (1)  1341.4  foot-tons ;  (2)  1883  foot-tons. 

Prob.  129.  A  through  Pratt  truss,  Fig.  60,  has  8  panels, 
each  18  feet  long  and  24  feet  deep :  find  the  maximum  chord 
stresses  due  to  a  passenger  locomotive  and  tender  followed 
by  a  uniform  train  load  of  3000  Ibs.  per  linear  foot. 

To  find  the  position  for  the  maximum  moment  in  2-4 
caused  by  the  live  load,  we  try  the  first  driver  at  the 
panel  point  4.  Then  for  the  uniform  train  load  we  have 
x  =  7  x  18  —  36  =  90  feet.  In  this  position,  the  total  live 
load  on  the  truss 

=  88  +  1.5  x  90  =  223  tons. 


172  ROOFS  AND  BRIDGES. 

Since  8  +  8  <  £  x  223,  and  8  +  8  +  20  >  |  x  223,  this  is 
the  correct  position  for  maximum  moment  in  2-4. 
For  this  load  the  reaction,  by  formula  (1),  is 

R  =  ^¥  (2388  +  88  x  90  +  .75  x  902)  =  113.8  tons  ; 
and  the  moment  at  3  is 

M  =  113.8  x  18  -  188  =  1860.4. 


. 
.-.  max.  stress  in  2-4  =       ~  =  77.5  tons  =  stress  in  4-6. 

Similarly,  the  maximum  stresses  in  the  other  three  chord 
members  are  found  to  be  the  following  : 

Max.  stress  in  3-5  =  -  128.3,  in  5-7  =  -  157.1, 
in  7-9  =  -  165.8  tons. 

NOTE.  —  To  determine  the  maximum  stress  in  7-9  put  10  feet  of  the 
train  to  the  left  of  the  point  10  ;  that  is,  put  the  third  tender  wheel  at 
8.  Then 

W=  88  +  1.5  x  82  =  211  tons. 

But  88  +  10  x  1.5  <  $  x  211;  therefore  this  is  about  the 
correct  loading. 

R  =  yff(2388  +  88  x  82  +  .75  x  S22)  =  101.7  tons. 
M  =  101.7  x  72  -  (2388  +  88  x  10  +  .75  x  102) 
=  3979.4  foot-tons. 

.-.  max.  stress  in  7-9  =  -  3979'4  =  -  165.8  tons. 

Prob.  130.  A  through  Pratt  truss,  Fig.  60,  has  8  panels, 
each  20  feet  long  and  24  feet  deep  :  find  (1)  the  maximum 
chord  stresses,  and  (2)  the  maximum  web  stresses,  in  each 
panel,  due  to  a  decapod  engine  and  tender  followed  by  a 
uniform  train  load  of  3000  Ibs.  per  linear  foot. 


BRIDGE  TRUSSES.  173 

(1)  To  find  the  position  for  maximum  stress  in  2-4  we 
try  the  2d  driver  at  4 ;  then  x  =  102.6  feet ;  and 

W=  112  +  1.5  X  102.6  =  266  tons. 

Since  |  x  266  =  33£  >  8  +  12.8,  and  <  8  +  12.8  +  12.8, 
this  is  the  correct  position  for  maximum  stress  in  2-4, 
although  if  the  third  driver  should  be  put  at  4  we  would 
get  about  the  same  stress,  as  8  +  12.8  -f  12.8  would  be  just 
less  than  £  of  the  load  on  the  truss.*  The  reaction  is 

R  =  _^  (2909.9  +  112  x  102.6  +  .75  x  ]M62)  =  139.3  tons. 

.-.  max.  stress  in  2-4  = 139'3  x  20  ~  152'4  =  109.7  tons 

^4 

=  stress  in  4-6. 

Similarly,  the  maximum  stresses  in  the  other  three  chord 
members  are  found  to  be  — 181.5,  —218.9,  and  —  226.4  tons. 

(2)  To  find  the  position  for  maximum  shear  in  2-3  put 
the  3d  driver  at  4 ;  then  x  =  106.8  feet,  and 

W=  112  +  1-5  x  106.8  =  272.2  tons. 

Since  8  + 12.8  + 12.8  <  %  x  272.2,  this  is  the  correct 
position  for  maximum  shear  in  2-3.  The  reaction  is 

R  =  ~  (2909.9  +  112  x  106.8  +  .75  x  106T82)  =  146.4  tons ; 

luU 

and  shear  in 

2-4  =  146.4  -  ^  (4.25  +  8.5)  - 1-  x  16.5  =  131.64  tons. 

.-.  max.  stress  in  2-3  =  - 131.64  x  1.302  =  -  171.4  tons. 

*  The  conditions  of  Arts.  47  and  48  may  sometimes  be  satisfied  by 
different  positions  of  the  load. 


174 


ROOFS  AND  BRIDGES. 


If  the  2d  driver  be  put  at  4  we  shall  obtain  about  the 
same  result. 

The  following  stresses  are  found  in  a  manner  similar  to 
the  above : 


Maximum  Stresses  in  the  Diagonals. 
2-3  =-171. 4  tons. 
3-6  =+130.7  tons. 
5-8  =  +  94.7  tons. 
7-10=+  63.6  tons. 
9-8'  =+  38. 3  tons. 
7'-6'  =  +  18.0  tons. 


Maximum  Stresses  in  the  Vertical*. 
3-4    =+51.1  tons. 
5_6    =-72. 7  tons. 
7-8    =-48. 9  tons. 
9-10  =  -  29.4  tons. 


Prob.  131.  A  through  Pratt  truss  has  7  panels,  each  20 
feet  long  and  20  feet  deep ;  the  dead  load  is  1600  Ibs.  per 
linear  foot,  the  live  load  is  a  passenger  locomotive  and 
tender  followed  by  a  uniform  train  load  of  3000  Ibs.  per 
Jinear  foot :  find  the  maximum  stresses  in  all  the  members. 


CHORD  STRESSES. 


MEMBERS. 

2-4,  4-6 

3-5 

5-7 

7-7' 

Dead  load  stresses    .... 
Live  load  stresses     .... 

+  48.0 
+  98.3 

-  80.0 
-157.3 

-  96.0 

-185.4 

-  96.0 
-185.4 

Maximum  stresses    .... 

+  146.3 

-237.3 

-281.4 

-281.4 

STRESSES  IN  THE  DIAGONALS. 


MEHBEBS. 

2-3 

3-6 

5-3 

7-S' 

7'-6' 

Dead  load  stresses  . 
Live  load  stresses  . 

-  67.9 
-138.9 

+  45.2 
+  98.7 

+22.6 
+64.6 

0.0 
+36.5 

0.0 
+16.7 

Maximum  stresses  . 

-206.8 

+  143.9 

+87.2 

+36.5 

-  5.9 

BRIDGE    TRUSSES. 


175 


The  live  load  stress  in  7'-6'  is  + 16.7,  while  the  dead 
load  stress  in  5-8'  or  5-8  is  +22.6.  Since  the  member 
7'— 6'  cannot  take  compression,  this  result  shows  that  the 
counter  7'-6'  is  not  needed  for  this  loading,  the  member 
5'-8'  taking  the  shear  as  tension. 

STRESSES  IN  THE  VERTICALS. 


MEMBERS. 

3-4 

5-6 

7-8 

Dead  load  stresses   

+  16.0 

-16.0 

-  0.0 

Live  load  stresses 

+39  6 

45  7 

25  8 

Maximum  stresses   

+  55  6 

-61  7 

25  8 

Prob.  132.  A  through  Pratt  truss,  Fig.  33,  has  10  panels, 
each  20  feet  long  and  20  feet  deep ;  the  dead  load  is  2000 
Ibs.  per  foot,  the  live  load  is  a  decapod  engine  and  tender 
followed  by  a  uniform  train  load  of  3000  Ibs.  per  foot :  find 
the  maximum  stresses  in  all  the  members. 

CHORD  STRESSES. 


MEMBERS. 

2-4,  4-6 

6-8 

8-10 

10-12 

9-11 

Dead  load  stresses  . 
Live  load  stresses    . 

+  90.0 
+  163.0 

+  160.0 

+280.8 

+210.0 
+357.7 

+  240.0 
+  398.0 

-250.0 
-406.6 

Maximum  stresses  . 

+253.0 

+440.6 

+  567.5 

+  638.0 

-656.6 

STRESSES  IN  THE  DIAGONALS. 


MEMBERS. 

2-3 

3-6 

5-8 

7-10 

9-12 

11-10' 

9'-S' 

Dead  load  stresses  .... 
Live  load  stresses  .   .   .  '  . 

-127.8 
-230.5 

+  98.9 
+186.5 

+  70.7 
+146.9 

+  42.4 
+111.4 

+14.1 
+80.2 

0.0 
+54.1 

0.0 
+32.3 

Maximum  stresses  .... 

-357.8 

+285.4 

+217.6 

+153.8 

+94.8 

+40.0 

-10.1 

176  ROOFS  AND  BRIDGES. 

STRESSES  IN  THE  VERTICALS. 


MEMBERS. 

3-4 

5-6 

7-3 

9-10 

11-12 

Dead  load  stresses  .  .  . 
Live  load  stresses    .  .  . 

+20.0 
+61.1 

-  50.0 
-103.9 

-  30.0 

-  78.8 

-10.0 
-56.7 

-  0.0 
-38.3 

Maximum  stresses  .  .  . 

+  71.1 

-153.9 

-108.8 

-66.7 

-38.3 

Since  the  live  load  stress  in  11-10'  or  11-10  is  +  54.1, 
while  the  dead  load  stress  in  the  same  diagonals  is  —  14.1, 
the  maximum  stress  is  the  difference,  or  +40.0.  Also, 
since  the  live  load  stress  in  9'-8'  or  9-8  is  +  33.1,  while  the 
dead  load  stress  is  —42.4,  the  maximum  stress  in  9'-8'  or 
9-8  is  —  9.3.  Hence  counters  are  needed  only  in  the  two 
middle  panels. 

Prob.  133.  A  through  Warren  truss  has  8  panels,  each  15 
feet  long  and  15  feet  deep ;  the  dead  load  is  2000  Ibs.  per 
linear  foot,  the  live  load  is  a  passenger  locomotive  and 
tender  followed  by  a  uniform  train  load  of  3000  Ibs.  per 
linear  foot :  find  the  maximum  stresses  in  all  the  members. 

Prob.  134.  A  deck  Pratt  truss  has  10  panels,  each  20  feet 
long  and  24  feet  deep ;  the  dead  load  is  given  by  formula  (1), 
Art.  15,  the  live  load  is  a  decapod  engine  and  tender  followed 
by  a  uniform  train  load  of  3000  Ibs.  per  linear  foot :  find 
the  maximum  stresses  in  all  the  members. 


CHAPTER   IV. 

MISCELLANEOUS  TRUSSES. 
KOOF  TRUSSES. 

Art.  50.  The  King  and  Queen  Truss  —  The  Fink 
Truss.  —  The  types  of  roof  trusses  most  commonly  used  for 
spans  from  30  feet  to  100  feet,  and  even  to  130  feet,  are 
shown  in  Chapter  I.  The  King  and  Queen  truss,  Fig.  7,  is 
a  common  form  for  ivooden  trusses,  or  for  trusses  that  are  all 
of  wood  except  the  verticals,  which  are  iron  or  steel  tie-rods. 
This  type  of  truss  is  sometimes  used  also  when  it  is  entirely 
made  of  steel. 

The  Belgian,  or  Fink  roof-truss,  Fig.  10,  is  a  very  common 
and  economical  type  of  truss  for  spans  up  to  130  feet.  The 
struts,  3-4,  5-6,  7-8,  are  normal  to  the  rafter,  dividing  it 
into  equal  parts,  and,  with  the  upper  chord,  are  made  either 
of  wood  or  iron.  The  other  members  are  ties,  and  are  made 
of  iron  or  steel.  This  type  of  truss  is  often  entirely  of  steel, 
and  is  commonly  used  for  iron  or  steel  roofs  over  mills, 
shops,  warehouses,  train-sheds,  etc.  Some  of  the  largest 
trusses  of  this  type  are  those  in  the  car-shops  of  the  Penn- 
sylvania railroad  at  Altoona,  Pa.,  having  a  span  of  132  feet. 

The  slope  of  the  rafter  is  usually  determined  by  the  kind 
of  roof  covering  used.  Slate  should  not  be  used  on  a  roof 
when  the  slope  is  less  than  1  vertical  to  3  horizontal,  and 
preferably  1  vertical  to  2  horizontal.  Gravel  should  not  be 
used  on  a  slope  greater  than  1  vertical  to  4  horizontal.  Tin 
177 


178  HOOFS  AND  BRIDGES. 

may  be  used  on  any  slope,  or  on  a  flat  roof.  Corrugated 
iron  should  not  be  used  on  a  slope  less  than  1  to  3;  for 
flatter  roofs  than  1  to  3,  of  corrugated  iron,  are  liable  to 
leak  under  a  driving  rain  as  the  usual  joints  are  not  tight. 
When  possible  the  slope  should  not  be  less  than  1  to  2. 

The  following  table  gives  the  approximate  weight  per 
square  foot  of  roof  coverings,  exclusive  of  steel  construction. 

APPROXIMATE  WEIGHT  PER  SQUARE  FOOT  OF  ROOF  COVERINGS. 

Corrugated  iron,  unbearded,  No.  26  to  No.  18 .     .     .     .  1  to  3  Ibs. 

Felt  and  asphalt,  without  sheathing 2  Ibs. 

Felt  and  gravel,  without  sheathing 8  to  10  Ibs. 

Slate,  without  sheathing,  fa"  to  \" 7  to  9  Ibs. 

Copper,  without  sheathing 1  to  \\  Ibs. 

Tin,  without  sheathing 1  to  l\  Ibs. 

Shingles,  with  lath 2J  Ibs. 

Skylight  of  glass,  T\"  to  £",  including  frame    ....  4  to  10  Ibs. 

White  pine  sheathing,  1"  thick "...  3  Ibs. 

Yellow  pine  sheathing,  1"  thick 4  Ibs. 

Spruce  sheathing,  1"  thick 2  Ibs. 

Lath  and  plaster  ceiling .  8  to  10  Ibs. 

Tile,  flat 16  to  20  Ibs. 

Tile,  corrugated •   .     .  *.  8  to  10  Ibs. 

Tile,  on  3"  fireproof  blocks 30  to  35  Ibs. 

The  weight  of  the  steel  roof  construction  must  be  added 
to  the  above.  For  ordinary  light  roofs  without  ceilings, 
the  weight  of  the  steel  construction  may  be  taken  at  5  Ibs. 
per  square  foot  for  spans  up  to  50  feet,  and  1  Ib.  additional 
for  each  10  feet  increase  of  span.  (Manual  of  useful  infor- 
mation and  tables  appertaining  to  the  use  of  Structural 
Steel,  as  manufactured  by  the  Passaic  Kolling  Mill  Co., 
Paterson,  N.  J.  By  George  H.  Blakeley,  C.E.) 

Prob.  135.  A  truss,  Fig.  62,  with  one  end  free,  has  its 
span  120  feet,  its  rise  30  feet,  the  ties,  3-4,  5-6,  7-8,  9-10, 


MISCELLANEOUS   TRUSSES. 


179 


11-12,  vertical,  dividing  the  rafter  into  6  equal  parts ;  the 
dead  load  per  panel  is  2.5  tons,  snow  load  per  panel  1.5  tons, 
normal  wind  load  per  panel,  wind  on  fixed  side,  2  tons :  find 


all  the  stresses  in  all  the  members  of  the  half  of  the  truss 
on  the  windward  side. 


STRESSES  IN  THE  LOWER  CHORD. 


MEMBERS. 

2-« 

6-S 

8-10 

10-12 

12-14 

Dead  load  stresses  .  .  . 
Snow  load  stresses  ,  .  . 
Wind  load  stresses  .  .  . 

+  27.5 
+  16.5 
+  17.8 

+25.0 
+  15.0 
+  15.6 

+22.5 
+  13.5 
+  13.4 

+20.0 
+  12.0 
+  11.1 

+  17.5 
+  10.5 
+  8.9 

Maximum  stresses  .  .  . 

+61.8 

+55.6 

+49.4 

+43.1 

+  36.9 

STRESSES  IN  THE  UPPER  CHORD. 


MEMBERS. 

2-3 

3-5 

5-7 

7-9 

9-11 

11-18 

Dead  load  stresses 
Snow  load  stresses 
Wind  load  stresses 

-30.75 
-18.45 
-14.52 

-27.95 
-16.77 
-13.00 

-25.16 
-15.09 
-11.48 

-22.35 
-13.41 
-  9.96 

-19.55 
-11.73 
-  8.44 

-16.75 
-10.05 
-  7.00 

Maximum  stresses 

-63.72 

-57.72 

-51.72 

-45.72 

-39.72 

-33.80 

180 


ROOFS  AND  BRIDGES. 
STRESSES  IN  THE  VERTICALS. 


MEMBERS. 

5-6 

7-8 

9-10 

11-12 

13-14 

Dead  load  stresses  .  . 
Snow  load  stresses  .  . 
Wind  load  stresses    . 

+  1.25 
+0.75 
+  1.12 

+  2.50 
+  1.50 

+2.24 

+3.75 
+2.25 
+  3.36 

+  5.00 
+  3.00 
+  4.46 

+  12.50 
+  7.50 
+  5.56 

Maximum  stresses  .  . 

+3.12 

+6.24 

+9.36 

+  12.46 

+  25.56 

(3-4  is  not  necessary  to  the  stability  of  the  truss. ) 
STRESSES  IN  THE  DIAGONALS. 


MEMBERS.                1       3-6 

5-8 

7-10 

C-12 

11-14 

Dead  load  stresses  . 
Snow  load  stresses  . 
Wind  load  stresses  . 

-2.80 
-1.68 
-2.48 

-3.52 
-2.11 
-3.16 

-  4.50 
-  2.70 
-  4.04 

-   5.55 
-  3.33 

-  5.00 

-  6.70 
-  4.02 
-  6.00 

Maximum  stresses  . 

-6.96 

-8.79 

-11.24 

-13.88  !  -16.72 

Prob.  136.  A  truss  of  the  type  of  Fig.  62,  with  one  end 
free,  has  its  span  150  feet,  its  rise  25  feet,  the  ties,  3—4, 
5-6,  etc.,  vertical,  dividing  the  rafter  into  six  equal  parts ; 
the  dead  load  per  panel  is  2.8  tons,  snow  load  per  panel 
1.5  tons,  normal  wind  load  per  panel,  wind  on  fixed  side, 
1.8  tons  :  find  all  the  stresses  in  all  the  members  of  the  half 
of  the  truss  on  the  windward  side. 

STRESSES  IN  THE  LOWER  CHORD. 


MEMBERS. 

2-6 

C-3 

8-10 

10-12 

12-14 

Dead  load  stresses  .  . 
Snow  load  stresses  .  . 
Wind  load  stresses  .  . 

+46.20 
+24.75 
+  22.79 

+42.00 
+22.50 
+  19.93 

+37.80 
+20.25 
+  17.06 

+33.60 
+  18.00 
+  14.20 

+  29.40 
+  15.75 
+  11.34 

Maximum  stresses  .  . 

+93.74 

+  84.43 

+  75.11 

+65.80 

+  56.49 

MISCELLANEO  US    TR  USSEH. 
STRESSES  IN  THE  UPPER  CHORD. 


181 


MEMBERS. 

2-3 

3-5 

5-7 

7-9 

9-11 

11-  3 

Dead  load  stresses  . 
Snow  load  stresses  . 
Wind  load  stresses. 

-48.97 
-26.24 
-20.74 

-44.52 

-23.85 
-18.32 

-40.07 
-21.47 
-15.91 

-35.62 
-19.08 
-  13.50 

-31.16 
-16.70 
-11.09 

-26.71 
-14.31 
-  8.96 

Maximum  stresses  . 

-95.95 

-86.69 

-77.45 

-68.20 

-58.95 

-49.98 

STRESSES  IN  THE  VERTICALS. 


MEMBERS. 

5-6 

7-, 

C-10 

11-12 

18-14 

Dead  load  stresses  .  .  . 
Snow  load  stresses  .  .  . 
Wind  load  stresses  .  .  . 

+  1.40 
+0.75 
+0.94 

+2.80 
+  1.50 
+  1.89 

+  4.20 

+2.25 
+2.84 

+   5.60 
+  3.00 

+  3.78 

+  17.02 
+  9.12 
+  5.76 

Maximum  stresses  .  .  . 

+3.09 

+6.19 

+9.29 

+  12.38 

+31.90 

STRESSES  IN  THE  DIAGONALS. 


MEMBERS. 

8-6 

5-9 

7-10 

9-12 

11-14 

Dead  load  stresses  .  .  . 
Snow  load  stresses  .  .  . 
Wind  load  stresses  .  .  . 

-  5.15 
-  2.76 
-  3.38 

-  5.60 
-  3.00 

-  3.82 

-  6.27 
-  3.36 

-  4.27 

-  7.28 
-  3.90 
-  4.97 

-  8.34 
-  4.47 
-  5.66 

Maximum  stresses  .  .  . 

-11.29 

-12.42 

-13.90 

-16.15 

-18.46 

Prob.  137.  A  Fink  truss,  Fig.  10,  with  one  end  free,  has 
its  span  150  feet,  its  rise  25  feet,  the  rafter  divided  into 
four  equal  parts  by  struts  drawn  normal  to  it,  the  dead  load 
per  panel  3  tons,  the 'snow  load  per  panel  1.5  tons,  the 
normal  wind  load  per  panel,  wind  on  fixed  side,  2  tons : 
find  all  the  stresses  in  all  the  members  of  the  half  of  the 
truss  on  the  windward  side. 


182 


ROOFS  AND  BRIDGES. 


STRESSES  IN  THE  LOWER  CHORD. 


MEMBERS. 

2-4 

4-6 

6-10 

Dead  load  stresses    .... 
Snow  load  stresses    .... 
Wind  load  stresses   .... 

+  31.50 
+  15.75 
+  15.84 

+  27.00 
+  13.50 
+  12.68 

+  18.00 
+  9.00 
+  6.34 

Maximum  stresses    .... 

+63.09 

+  53.18 

+  33.34 

STRESSES  IN  THE  UPPER  CHORD. 


MEMBERS. 

2-3 

3-5 

6-7 

7-9 

Dead  load  stresses   .... 
Snow  load  stresses  .... 
Wind  load  stresses  .... 

-33.16 
-16.58 
-14.36 

-32.22 
-16.11 
-14.36 

-31.29 
-15.65 
-14.36 

-30.36 
-15.18 
-14.36 

Maximum  stresses  .... 

-64.09 

-62.69 

-61.30 

-59.90 

STRESSES  IN  THE  WEB  MEMBERS. 


MEMBERS. 

3-4,7-8 

5-6 

4^5,5-8 

6-3 

8-9 

Dead  load  stresses  .  .  . 
Snow  load  stresses  .  .  . 
Wind  load  stresses  .  .  . 

-2.85 
-1.43 
-2.00 

-  5.70 
-  2.85 
-  4.00 

+4.50 
+  2.25 
+3.12 

+  9.00 
+  4.50 
+  6.32 

+  13.50 
+  6.75 
+  9.48 

Maximum  stresses  .     . 

-6.28 

-12.55 

+9.87 

+  19.82 

+29.73 

Art.  51.    The  Crescent  Truss,  Figs.  63  and  64,  is  a 

good  form  for  comparatively  large  spans.     Riveted  iron- 
work is  used  throughout. 

Prdb.  138.  A  circular  Crescent  truss,  Fig.  63,  with  one 
end  free,  has  its  span  160  feet,  the  rise  of  the  upper  chord 
32  feet,  the  rise  of  the  lower  chord  20  feet,  the  number  of 


MISCELLANEOUS    TRUSSES. 


183 


panels  8,  the  apexes  of  both  the  upper  and  lower  chords 
lying  on  arcs  of  circles,  dividing  them  into  8  equal  parts ; 
the  dead  load  is  3  tons  per  panel,  the  wind  load  is  2  tons 


per  panel,  the  wind  being  supposed  to  act  vertically  on  the 
fixed  side  only : l  find  all  the  stresses  in  all  the  members  of 
the  half  of  the  truss  on  the  windward  side. 

The  computation  of  the  stresses  is  effected  by  the  appli- 
cation of  the  principles  of  Chapter  I. 

The  radius  of  the  upper  chord  =  116  feet. 
The  radius  of  the  lower  chord  =  170  feet. 
The  lengths  of  2-3,  3-5,  etc.     =  22.04  feet. 
The  lengths  of  2-4,  4-6,  etc.     =  20.81  feet. 

The  horizontal  distance  of  each  apex  from  the  left  abut- 
ment 2,  and  its  vertical  distance  above  the  line  2-2',  are  as 
follows : 

(3)  (5)        (7)          (9) 
17.33,    36.94,    58.06,    80  feet. 
13.61,    23.70,    29.91,    32  feet. 

(4)  (6)        (8)         (10) 
18.93,    38.76,    59.22,    80  feet. 

8.65,    14.92,    18.72,    20  feet. 


Hor.  distance  from  2, 
Height  above  line  2-2', 

Hor.  distance  from  2, 
Height  above  line  2-2', 


1  If  the  normal  wind  pressure  were  taken,  it  would  have  a  different 
value  for  each  panel,  owing  to  the  curved  surface  of  the  rafters,  which 
would  increase  the  difficulty  of  the  computation. 


184 


HOOFS  AND  BRIDGES. 


The  dead  load  reaction  =  10.5  tons. 

The  stresses  in  the  chords  are  best  found  by  moments ; 
for  the  other  members  the  method  of  resolution  of  forces 
may  be  used. 

Thus,  to  compute  the  dead  load  stress  in  2-3,  we  find  the 
lever  arm  of  2-3  to  be  4.89  feet. 

.-.  dead  load  stress  in  2-3  =  - 10-5  x  18-93  =  _  40.7  tons. 

4.89 

Similarly,  the  stress  in  2^  =  10'5  x  ^7'33  =  35.1  tons. 

5.18 

To  find  the  stresses  in  3-4  and  3-5  pass  a  section  cutting 
2-4,  3-4,  and  3-5.  Then,  denoting  the  stresses  in  3-4  and 
3-5  by  sv  and  s^,  we  have,  for  horizontal  and  vertical  com- 
ponents respectively, 

35.1  x  .9095  +         1'6_—  Sl  +  .889  s2  =  0, 
^Di2  +  4.962 

and       35.1  x  .4156  - 1||  s,  +  .4579  s2  +  7.5  =  0. 

.-.  Si  =  +    5.0  tons  =  stress  in  3-4 ; 
and  s2  =  —  37.7  tons  =  stress  in  3-5. 

The  wind  load  reaction  =  5.1  tons. 

A  wind  load  stress  in  2-3  =  -  5>1  *  18-93  =  _  19.7  tons. 

4.89 

In  this  way  all  the  stresses  may  be  found. 
STRESSES  IN  THE  TOP  CHORD. 


MEMBERS. 

2-8 

3-6 

5-7 

7-9 

Dead  load  stresses  .  .  . 
Wind  load  stresses  .  .  . 

-40.7 
-19.7 

-37.7 
-17.0 

-37.8 
-16.6 

-37.8 
-14.8 

Maximum  stresses  .  .  . 

-60.4 

-64.7 

-54.4 

-62.6 

MISCELLANEO  US   TR  USSES. 


185 


STRESSES  IN  THE  BOTTOM  CHORD. 


MEMBERS. 

2-4 

4-6 

6-8 

8-10 

Dead  load  stresses  .  .  . 
Wind  load  stresses  .  .  . 

+36.1 
+  16.0 

+  36.9 
+  16.4 

+37.8 
+15.0 

+38.1 
+  12.7 

Maximum  stresses  .  .  . 

+  51.1 

+  53.3 

+  52.8 

+  50.8 

STRESSES  IN  THE  WEB  MEMBERS. 


MEMBERS. 

3-4 

5-6 

7-8 

9-10 

4-5 

6-7 

8-9 

Dead  load  stresses 
Wind  load  stresses 

+  5.0 
+2.2 

-1.8 

+2.0 

+  5.0 

+  2.8 

-1.4 

+  1.4 

+5.1 
-2.0 

-0.6 
-2.0 

+4.2 
+2.4 

Max.  compression 
Max.  tension  .  .  . 

+0.0 

+7.2 

-1.8 

+0.2 

+  0.0 

+  7.8 

-1.4 

+0.0 

+0.0 
+  5.1 

-2.6 
-0.0 

+  0.0 
+6.6 

Prob.  139.    A  circular  Crescent  truss,  Fig.  64,  with  one 
end  free,  has  its  span  160  feet,  the  rise  of  the  upper  chord 


32  feet,  the  rise  of  the  lower  chord  20  feet,  the  apexes  of 
both  the  upper  and  lower  chords  lying  on  arcs  of  circles. 
The  upper  chord  has  8  panels,  all  of  the  same  length.  The 
lower  chord  has  7  panels ;  each  of  the  5  panels  4-6.  6-8, 


186 


ROOFS  AND  BRIDGES. 


etc.,  is  equal  to  one-eighth  of  the  lower  chord,  while  the  two 
end  panels,  2-4  and  2'-4',  are  each  equal  to  1^  of  the  other 
panels,  that  is,  equal  to  1^  eighths  of  the  lower  chord ;  the 
dead  load  is  3  tons  per  panel  and  the  wind  load  is  2  tons 
per  panel,  the  wind  being  supposed  to  act  vertically  on  the 
fixed  side :  find  all  the  stresses  in  all  the  members  of  the 
half  of  the  truss  on  the  windward  side. 

Here  the  radii  of  the  upper  and  lower  chords  are  116  and 
170  feet,  respectively,  as  in  Prob.  138,  and  the  coordinates 
of  the  apexes  of  the  upper  chord  are  also  the  same  as  in 
Prob.  138. 

The  length  of  2-4  is  found  to  be  31.21  feet,  and  the  coor- 
dinates of  4  to  be  28.77  feet  and  12.09  feet. 

The  reactions  are  the  same  as  in  Prob.  138. 

.-.  Dead  load  stress  in  2^3  =  - 10-5  x  28-77  =  _  36.6  tons. 

8.26 

Dead  load  stress  in  2-4  =     10-5  xj-7-33  =     31.2  tons. 

5.83 

Wind  load  stress  in  2-3  =  -  51  *  Jj.8'77  =  -  17.7  tons. 


Wind  load  stress  in  2-4  =      B*  *  17'33  =     15.1  tons. 

5.83 

STRESSES  IN  THE  TOP  CHORD. 


MEMBERS. 

2-3 

3-5 

5-7 

7-9 

Dead  load  stresses  

-36.6 

—40  5 

-393 

-38  7 

17  7 

18  4 

16  8 

14  4 

Maximum  stresses  

-64  3 

-589 

-56  1 

53  1 

MISCELLANEOUS   TRUSSES,  187 

STRESSES  IN  THE  BOTTOM  CHORD. 


MEMBERS. 

2-4 

4-6 

6-S 

8-3' 

]  )ead  load  stresses  
Wind  load  stresses  

+  31.2 
+  15  1 

+  35.1 
+  16  2 

+  36.6 

+  14  7 

+  36.6 

+  12  4 

Maximum  stresses  

+46.8 

+51.3 

+  51.3 

+  49.0 

STRESSES  IN  THE  WEB  MEMBERS. 


MEMBERS. 

8-1 

4-5 

5-8 

e-7 

7-S 

S-9 

Dead  load  stresses    .  . 
Wind  load  stresses   .  . 

+  7.2 
+  3.0 

+3.2 
+  1.4 

+4.1 

+  2.2 

+2.7 
+  2.6 

+  3.1 
+2.9 

+  3.0 

+2.7 

Maximum  stresses  .  .  . 

+  10.2 

+4.6 

+  6.3 

+  5.3 

+  6.0 

+  5.7 

BRIDGE  TRUSSES. 

Art.  52.  The  Pegram  Truss  —  The  Parabolic  Bow- 
string Truss.  —  The  Pegram  truss,  Fig.  65,  consists  of  the 
same  number  of  panels  in  each  chord.  All  the  panels  of 


each  chord  are  of  equal  length,  the  upper  chord  panels 
being  shorter  than  the  lower.  The  apexes  of  the  upper 
chord  lie  on  an  arc  of  a  circle,  the  chord  of  which,  3-3',  is 
about  one  and  one  half  panel  lengths  shorter  than  the  span. 
The  rise  of  the  upper  chord  is  such  as  to  make  the  posts, 


188  ROOFS  AND  BRIDGES. 

4-5,  6-7,  8-9,  nearly  equal  in  length,  or  it  may  be  so  taken 
that  the  posts  will  decrease  in  length  toward  the  ends.  In 
a  deck  bridge  the  upper  chord  is  made  straight  and  the 
lower  chord  curved. 

Prob.  140.  A  through  Pegram  truss,  Fig.  65,  has  a  span 
of  200  feet,  divided  into  7  panels,  each  28.57  feet  long ;  the 
upper  apexes  lie  on  an  arc  of  a  circle,  the  center  height 
being  15  feet  above  the  chord  3-3',  which  is  160  feet  long ; 
the  upper  panels  are  all  of  equal  length,  the  members  2-3, 
4-5,  6-7,  8-9  are  struts,  all  the  remaining  web-members  are 
ties,  the  dead  and  live  loads  are  10  tons  and  18  tons  per 
panel  per  truss :  find  the  stresses  in  all  the  members. 

Here  the  radius  of  the  upper  chord  =  220.83  feet. 

The  lengths  of  the  upper  panels  3-5,  5-7,  etc. =23.37  feet. 

The  horizontal  distance  of  each  upper  apex  from  the  left 
abutment  2,  and  its  vertical  height  above  the  lower  chord 
2-2',  are  the  following : 

(3)        (5)         (7)        (9) 

Hor.  distance  from  2,  20.00,  42.20,  65.05,  88.05. 

Height  above  lower  chord,  24.00,  31.29,  36.23,  38.65. 

The  dead  load  reaction  =  30  tons. 
Then,  dead  load  stress  in 
30x  42.20- 


31.29 
Also,  dead  load  stress  in  9-9' 

30  x  4  x  28.57  -  60  x  28.57 
38.65 

Similarly,  dead  load  stress  in  3-5 
30  x  28.57 
25.47 


-  44.3  tons. 


MISCELLANEOUS   TRUSSES.  189 

To  find  the  dead  load  stress  in  any  web  member,  as  5-6, 
pass  a  section  cutting  5-7,  5-6,  and  4-6,  and  take  moments 
around  the  intersection  of  5-7  and  4-6,  which  is  102.53  feet 
to  the  left  of  2. 


.-.  dead  load  stress  in  5-6 

^30x102.53-10x131.1 
144.1 


=  12.2  tons. 


The  stresses  in  the  other  web  members  are  found  in  like 
manner. 

The  live  load  stresses  in  the  chords  and  in  2-3  and  3-4 
are  a  maximum  for  a  full  load,  and  may  therefore  be 
found  by  multiplying  the  corresponding  dead  load  stresses 
by  1.8. 

For  a  maximum  in  5-6  and  4-5  the  live  load  covers  all  the 
joints  from  the  right  to  6.  The  reaction  for  this  loading  is 


.-.  live  load  stress  in  5-6  =  18  *  15  x       ~  =  27.4  tons. 


For  a  maximum  in  9-8'  the  joints  4',  6',  and  8'  are  loaded. 
The  reaction  for  this  loading  is 

^  =  ^(1+2  +  3)  =  !^  tons, 

which  is  also  the  maximum  shear  in  this  panel.  Since  the 
upper  and  lower  chord  members  of  this  panel  are  horizontal, 
the  stress  in  9-8'  is  found  by  Art.  18. 


.-.  max.  stress  in  9-8'  =        x  =  18-5  tons- 

7       38.65 


190 


ROOFS  AND   BRIDGES. 
STRESSES  IN  THE  UPPER  CHORD. 


MEMBERS. 

8-5 

5-T 

7-9 

9-9' 

Dead  load  stresses    

—33.6 
-60.4 

-  42.4 

-  76.2 

-  45.1 
-  81  1 

-  44.3 

-  79  7 

Maximum  stresses   

-94.0 

-118.6 

-126.2 

-124.0 

STRESSES  IN  THE  LOWER  CHORD. 


MEMBERS. 

2-4 

4-6 

C-3 

8-8' 

Dead  load  stresses   

+25.0 
+  45  0 

+  36.1 
+  65  0 

+  41.6 
+  74  9 

+  44.6 
+  80  3 

+  70  0 

+  101  1 

•  +H6  5 

+  124  9 

STRESSES  IN  THE  WEB  TIES. 


MEMBERS. 

3-4 

£-6 

7-3 

9-3' 

9-->r 

Dead  load  stresses  .  . 
Live  load  stresses    .  . 

+20.6 
+37.2 

+  12.2 

+27.4 

+  6.2 

+22.7 

+  0.0 
+  18.5 

+  0.0 
+  14.1 

Maximum  stresses  .  . 

+  57.8 

+39.6 

+28.9 

+  18,5 

+  14.1 

STRESSES  IN  THE  WEB  STRUTS. 


MEMBERS. 

2-3 

4-5 

6-7 

8-9 

Dead  load  stresses    .  . 
Live  load  stresses  .  .  . 

-  38.8 
-  69.8 

-10.6 

-27.8 

-   1.3 
-17.3 

-  0.0 
-12.0 

Maximum  stresses    .   . 

-108.6 

-38.4 

-18.6 

-12.0 

MISCELLANEOUS  TBUSSES. 


191 


See  Framed  Structures,  by  Johnson,  Bryan,  and  Turneaure ;  also 
Engineering  News,  Dec.  10  and  17,  1887,  and  Feb.  14,  1891,  where  the 
answers  differ  very  slightly  from  the  above  on  account  of  the  center 
member  9-9'  being  ^  less. 

Prob.  141.  A  through  parabolic  bowstring  truss,  Fig.  66, 
has  8  panels,  each  24  feet  long,  and  32  feet  center  depth, 

9 r 

f 


U  6  8  10  8'  6'  K'  2' 

mig.ee 

the  verticals  are  ties  and  the  diagonals  are  struts ;  the  dead 
and  live  loads  are  10  tons  and  15  tons  per  panel  per  truss : 
find  the  stresses  in  all  the  members. 

STRESSES  IN  THE  TOP  CHORD. 


MEMBERS. 

2-3 

3-5 

5-7 

7-9 

Dead  load  stresses    .... 
Live  load  stresses  

-  69.4 
-104.1 

-  65.0 
-  97.5 

-  61.8 
-  92.7 

-  60.2 
-  90.3 

Maximum  stresses    .... 

-173.5 

-162.5 

-154.5 

-150.5 

Dead  load  stress  in  each  panel  of  lower  chord  =  -f  60  tons. 
Live  load  stress  in  each  panel  of  lower  chord  =  +  90  tons. 
Maximum  stress  in  each  panel  of  lower  chord  =  -f  150  tons. 

STRESSES  IN  THE  DIAGONALS. 


MEMBERS. 

3-6 

5-S 

7-10 

4-5 

6-7 

8-9 

Dead  load  stresses  . 
Live  load  stresses  . 

0.0 
-13.0 

0.0 
-15.9 

0.0 
-18.0 

0.0 
-15.9 

0.0 
-18.0 

0.0 
-18.1 

Maximum  stresses. 

-13.0 

-15.9 

-18.0 

-15.9 

-18.0 

-18.1 

192  ROOFS  AND  BRIDGES. 

STRESSES  IN  THE  VERTICALS. 


MEMBERS. 

3-4 

6-6 

7-3 

9-10 

Dead  load  stresses  .... 
Live  load  stresses    .... 

+  10.0 
+  15.0 

+  10.0 
+  19.7 

+  10.0 

+  22.5 

+  10.0 
+23.4 

Maximum  stresses  .... 

+25.0 

+  29.7 

+32.5 

+33.4 

Art.  53.  Skew  Bridges  are  those  in  which  the  end 
supports  of  one  truss  are  not  directly  opposite  to  those  of 
the  other.  The  trusses  of  a  skew  bridge  are  usually  placed 
so  that  the  intermediate  panel  points  are  directly  opposite 
in  the  two  trusses,  and  the  floor  beams  are  at  right  angles 
to  the  trusses.  When  the  skew  is  the  same  at  each  end  the 
trusses  are  symmetrical;  otherwise  they  are  unsymmetrical. 
In  the  analysis  of  unsymmetrical  trusses,  each  truss  must 
be  treated  separately ;  and  the  stresses  are  to  be  computed 
for  all  the  members  of  the  truss. 

Prob.  142.  Fig.  68  is  a  plan  and  Fig.  67  is  the  elevation 
of  one  of  the  two  trusses  of  an  unsymmetrical  through 


t 4 s 8 

.18        "16  'H  1> 


• 


Wie-  68 


MISCELLANEOUS   TRUSSES.  193 

Pratt  bridge;  the  span  2-18  is  120  feet,  the  depth  is  20 
feet,  the  panels  3-5  and  2-4  are  each  18  feet,  the  panels 
13-15  and  16-18  are  each  12  feet,  the  other  panels  are  each 
15  feet,  the  inclination  of  the  end  posts  2-3  and  15-18  is 
the  same  as  that  of  the  diagonals  5-8,  7-10,  10-11,  12-13, 
etc.,  and  the  inclination  of  the  two  hip  verticals  3-4  and 
15-16  is  the  same;  the  dead  and  live  loads  are  1000  Ibs.  and 
2000  Ibs.  per  foot  per  truss :  find  all  the  stresses  in  all  the 
members. 

\Ye  first  find  the  chord  stresses  due  to  the  dead  load  as 
follows : 

Dead  panel  load  at     4  =  \  (9  +  7.5)  =  8.25  tons. 
Dead  panel  load  at  16  =  %  (6  +  7.5)  =  6.75  tons. 
Dead  panel  load  at  6,  8,  10,  12,  14  =  7.50  tons. 
Dead  load  reaction  at    2  =  25.5  tons. 
Dead  load  reaction  at  18  =  27.0  tons. 

tan  Z  3-2      with  vertical  =  .75  =  tan  Z  15-18  with  vertical, 
tan  Z  3-4      with  vertical  =  .15  =  tan  Z  15-16  with  vertical, 
tan  Z  3-6      with  vertical  =  .9. 
tan  Z 14-15  with  vertical  =  .6. 

Hence  by  (2)  of  Art.  19  we  have  the  following  dead  load 
stresses : 

Dead  load  stress  in  2-4      =  25.5  x    .75  =  19.1  tons. 

Dead  load  stress  in  4-6      =  19.1  +  8.25  x  .15  =  20.3  tons. 
Dead  load  stress  in  16-18  =  27.0  x    .75  =  20.3  tons. 

Dead  load  stress  in  14-16  =  20.3  -  6.75  x  .15  =  19.3  tons, 
etc.  etc.  etc. 


194 


EOOFS  AND  BRIDGES. 


Also  the  following  greatest  live  load  stresses : 

Stress  in  3-6  =  ^  [13.5  x  12  +  15  x  285]  1.345 

=  +  49.7  tons. 

Stress  in      5-6  =  -  ^  [162  + 15  x  198]  =  -  26.1  tons. 
Stress  in  12-13  =  ^  [297  + 15  x  222]  1.166  =  +  35.2  tons. 

CHORD  STRESSES. 


MEMBERS. 

2-4 

4-6 

6-8 

8-10 

10-12 

12-14 

14-16 

16-18 

7-11 

Dead  load  stresses  .  .  . 
Live  load  stresses    .  .  . 

+19.1 
+38.2 

+20.3 
+40.6 

+  85.8 
+  71.6 

+  48.1 

+  86.2 

+  41.0 

+  82.0 

+31.5 
+63.0 

+19.3 
+38.6 

+20.3 
+40.6 

-  44.8 

-  89.6 

Maximum  stresses  .  .  . 

+57.3 

+60.9 

+107.4 

+129.8 

+123.0 

+94.5 

+57.9 

+60.9 

-134.4 

STRESSES  IN  THE  DIAGONALS. 


MEMBERS. 

3-4 

3-6 

5-8 

7-10 

9-12 

11-14 

Dead  load  stresses  . 
Live  load  stresses  . 

+  8.3 
+  16.6 

+23.2 
+49.7 

+  12.2 
+35.1 

+  2.8 
+  23.0 

-  6.6 
+  13.5 

-15.9 
+  6.3 

Maximum  stresses 

+24.9 

+  72.9 

+47.3 

+25.8 

+  6.9 

-  9.6 

MEMBERS. 

15-16 

14-15 

12-13 

10-11 

8-9 

6-7 

Dead  load  stresses  . 
Live  load  stresses  . 

+  6.8 
+  13.6 

+23.6 

+48.8 

+  15.9 
+35.2 

+  6.6 
+23.9 

-  2.8 
+  14.7 

-12.2 

+  7.7 

Maximum  stresses  . 

+20.4 

+  72.4 

+  51.1 

+30.5 

+  11.9 

-  4.5 

We  see  from  the  above  that,  theoretically,  the  diagonals 
6  -7  and  11-14  are  not  needed. 


MISCELLANEOUS   TRUSSES. 
STRESSES  IN  THE  POSTS. 


195 


MEMBERS. 

2-3 

5-6 

7-8 

9-10 

11-12 

11-14 

15-18 

Dead  load  stresses  . 
Live  load  stresses  . 

-31.9 
-63.8 

-  9.8 
-26.1 

-  2.3 
-17.1 

-  0.0 
-12.6 

-  6.3 
-20.5 

-12.8 
-30.2 

-  33.8 
-  67.6 

Maximum  stresses  . 

-95.7 

-35.9 

-19.4 

-12.6 

-25.8 

-43.0 

-101.4 

Prob.  143.  Let  the  dimensions  of  Fig.  67  be  as  follows : 
span  =  144  feet,  depth  =  24  feet,  the  panels  3-5  and  2-4 
=  21  feet,  the  panels  13-15  and  16-18  =  15  feet,  all  the 
other  panels  =  18  feet;  let  the  dead  and  live  loads  be 
800  Ibs.  and  2000  Ibs.  per  foot  per  truss:  find  all  the 
stresses  in  all  the  members. 


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STANDARD  TEXT  BOOKS.  5 

GOODEVE,  T.  M.    A  Text-Book  on  the  Steam-Engine. 

With  a  Supplement  on  Gas-Engines.  Twelfth  Edition,  enlarged. 
143  illustrations.  12mo,  cloth $2.00 

GUY,    A.    E.     Experiments    on    the    Flexure    of    Beams, 

resulting  in  the  Discovery  of  New  Laws  of  Failure  by  Buckling. 
Reprinted  from  the  "American  Machinist."  With  diagrams  and 
folding  plates.  8vo,  cloth,  illustrated net,  $1 .25 

HAEDER,  HERMAN,  C.E.  A  Handbook  on  the  Steam- 
Engine.  With  especial  reference  to  small  and  medium-sized  en- 
gines. Third  English  Edition,  re-edited  by  the  author  from  the 
second  German  edition,  and  translated  with  considerable  addi- 
tions and  alterations  by  H.  H.  P.  Powles.  12mo,  cloth.  Nearly 
1100  illustrations $3 .00 

HALL,  WM.  S.,  Prof.     Elements  of  the  Differential  and 

Integral  Calculus.     Sixth  Edition,  revised.    8vo,  cloth,  illustrated. 

net,  $2.25 

Descriptive  Geometry ;   with  Numerous  Problems  and 

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a  4to  Atlas  of  illustrated  problems.  Two  vols.,  cloth.  .  .net,  $3 .50 

HALSEY,  F.  A.     Slide-Valve  Gears :    an  Explanation  of  the 

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trated. Eighth  Edition,  revised  and  enlarged.  12r»«,  cloth.  $1 .50 

HANCOCK,    HERBERT.     Text-Book    of    Mechanics    and 

Hydrostatics.     With  over  500  diagrams.     8vo,  cloth $1.75 

HARDY,    E.     Elementary   Principles   of   Graphic   Statics. 

Containing  192  diagrams.     8vo,  cloth,  illustrated SI  .50 

HAY,  A.     Alternating  Currents ;   Their  Theory,  Generation 

and  Transformation.     8vo,  cloth,  illustrated net,  $2.50 

HECK,  Prof.  R.  C.  H.     The  Steam-Engine.     Vol.  I.     The 

Thermodynamics  and  the  Mechanics  of  the  Engine.  8vo,  cloth, 
391  pp.,  illustrated net,  $3.50 

—  Vol.    II.  Form,    Construction,    and   Working   of   the 

Engine.     The  Steam-Turbine In  Press. 

HERRMANN,  GUSTAV.  The  Graphical  Statics  of  Mech- 
anism. A  Guide  for  the  Use  of  Machinists,  Architects,  and 
Engineers;  and  also  a  Text-Book  for  Technical  Schools.  Trans- 
lated and  annotated  by  A.  P.  Smith,  M.E.  7  folding  plates. 
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HIROI,    I.     Statically-Indeterminate    Stresses   in    Frames 

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ture Room  and  Laboratory  Manual  for  Colleges  and  Universities, 
With  a  bibliography  and  130  figures  and  diagrams.  300  pages. 

't,  83  . 


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HUTCHINSON,  R.  W.,  Jr.     Long-Distance  Electric  Power 

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bution. 12mo,  cloth,  illustrated  ......................  In  Press. 

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tion, and  Maintenance  of  Electrical  Mining  Machinery.  12mo, 
cloth,  illustrated  ....................................  In  Press. 

JAMIESON,  ANDREW,  C.E.  A  Text-Book  on  Steam  and 
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Art,  City  and  Guilds  of  London  Institute,  and  other  Engineering 
Students.  Fourteenth  Edition,  revised.  Illustrated.  12mo,  cloth. 

$3.00 

--  Elementary  Manual  on  Steam,  and  the  Steam-Engine  . 
Specially  arranged  for  the  Use  of  First-  Year  Science  and  Art, 
City  and  Guilds  of  London  Institute,  and  other  Elementary 
Engineering  Students.  Tenth  Edition,  revised.  12mo,  cloth.  $1.50 

JANNETTAZ,  EDWARD.     A  Guide  to  the  Determination 

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Brooklyn  Polytechnic  Institute.  12mo,  cloth  ............  $1  .50 

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STANDARD  TEXT  BOOKS.  7 

KEMP,  JAMES  FURMAN,   A.B.,   E.M.     A  Handbook   of 

Rocks;  for  use  without  the  microscope.  With  a  glossary  of  the 
names  of  rocks  and  other  lithological  terms.  Second  Edition, 
revised.  8vo,  cloth,  illustrated $1 .50 

KLEIN    J.    F.     Design    of    a    High-Speed    Steam-engine. 

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KNIGHT,  A.  M.,  Lieut.-Com.  U.S.N.  Modern  Seaman- 
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cloth,  illustrated.     Third  Edition,  revised net,  $6 . 00 

Half  morocco $7 . 50 

KNOTT,  C.  G.,  and  MACKAY,  J.  S.     Practical  Mathematics. 

With  numerous  examples,  figures,  and  diagrams.  New  Edition. 
8vo,  cloth,  illustrated $2 .00 

KRAUCH,    C.,    Dr.     Testing    of    Chemical    Reagents    for 

Purity.  Authorized  translation  of  the  Third  Edition,  by  J.  A. 
Williamson  and  L.  W.  Dupre.  With  additions  and  emendations 
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LASSAR-COHN,  Dr.  An  Introduction  to  Modern  Scien- 
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sity Extension  Students  and  general  readers.  Translated  from 
the  author's  corrected  proofs  for  the  second  German  edition  by 
M.  M.  Pattison  Muir,  M.A.  12mo,  cloth,  illustrated $2.00 

LODGE,    OLIVER    J.     Elementary   Mechanics,    including 

Hydrostatics  and  Pneumatics.     Revised  Edition.     12mo,  cloth. 

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LUCKE,   C.   E.     Gas  Engine   Design.     With  figures  and 

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LUQUER,  LEA  McILVAINE,  Ph.D.  Minerals  in  Rock  Sec- 
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in  Technical  and  Scientific  Schools.  Nev)  Edition,  revised.  8vo, 
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MARKS,  G.  C.  Hydraulic  Power  Engineering.  A  Prac- 
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MERCK,     E.     Chemical    Reagents;     Their    Purity    and 

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MILLER,  E.  H.  Quantitative  Analysis  for  Mining  Engi- 
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MINIFIE,   WM.     Mechanical  Drawing.     A  Text-Book   of 

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plained; the  Practical  Problems  are  arranged  from  the  most 
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chinery: an  Introduction  to  Isometrical  Drawing,  and  an  Essay 
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Ninth  Edition.  12mo,  cloth $2.00 

MOSES,  ALFRED  J.,  and  PARSONS,  C.  L.     Elements  of 

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Practical  Standpoint.  336  illustrations.  New  and  enlarged 
Edition.  8vo,  cloth $2 . 50 

MOSS,  S.  A.  Elements  of  Gas-Engine  Design.  Reprint 
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at  Cornell  University  in  1902.  16mo,  cloth,  illustrated.  (Van 
Nostrand's  Science  Series.) $0 .50 

NASMITH,  JOSEPH.  The  Student's  Cotton  Spinning. 
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NIPHER,  F.  E.,  A.M.     Theory  of  Magnetic  Measurements. 

With  an  Appendix  on  the  Method  of  Least  Squares.  12mo, 
cloth $1 .00 

NUGENT,   E.     Treatise   on   Optics;    or,   Light   and  Sight 

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STANDARD  TEXT  BOOKS.  9 

OLSEN,  Prof.  J.  C.    Text-Book  of  Quantitative  Chemical 

Analysis  by  Gravimetric,  Electrolytic,  Volumetric,  and  Gaso- 
metric  Methods.  With  seventy-two  Laboratory  Exercises  giving 
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OUDIN,  MAURICE  A.     Standard  Polyphase  Apparatus  and 

Systems.  With  many  diagrams  and  figures.  Third  Edition, 
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PALAZ,  A.,  Sc.D.     A  Treatise  on  Industrial  Photometry, 

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tion from  the  French  by  George  W.  Patterson,  Jr.  8vo,  cloth, 
illustrated $4.00 

PARSHALL,  H.  F.,  and  HOBART,  H.  M.  Armature  Wind- 
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PATTON,     HT^B.     Lej^fl^JJoteaXon     Crystallography. 

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PAULDING,  CHAS.  P.  Practical  Laws  and  Data  on  Con- 
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The  Transmission  of  Heat  through  Cold-storage  In- 
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PERRINE,  F.  A.  C.,  A.M.,  D.Sc.     Conductors  for  Electrical 

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PERRY,  JOHN.     Applied  Mechanics.     A  Treatise  for  the 

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« 

PLATTNER.  Manual  of  Qualitative  and  Quantitative  Analy- 
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and  enlarged,  by  Prof.  Th.  Richter,  of  the  Royal  Saxon  Mining 
Academy.  Translated  by  Prof.  H.  B.  Cornwall,  assisted  by 
John  H.  Caswell.  Illustrated  with  78  woodcuts.  Eighth  Edition, 
revised.  463  pages.  8vo,  cloth net,  $4.00 

POPE,  F.  L.  Modern  Practice  of  the  Electric  Telegraph. 
A  Technical  Handbook  for  Electricians,  Managers,  and  Operators. 
Seventeenth  Edition,  rewritten  and  enlarged,  and  fully  illustrated. 
8vo,  cloth $1 . 50 

PRELINI,   CHARLES.     Tunneling.    A  Practical   Treatise 

containing  149  Working  Drawings  and  Figures.  With  additions 
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Third  Edition,  revised.  8vo,  cloth,  illustrated $3 .00 

Earth  and  Rock  Excavation.  A  Manual  for  Engi- 
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trated. 350  pp net,  $3.00 

Retaining  Walls  and  Dams.     8vo,  cloth,  illustrated. 

In  Press. 

PRESCOTT,  Prof.  A.  B.     Organic  Analysis.    A  Manual  of 

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Compounds  in  Common  Use;  a  Guide  in  the  Qualitative  and 
Quantitative  Analysis  of  Organic  Materials  in  Commercial  and 
Pharmaceutical  Assays,  in  the  Estimation  of  Impurities  under 
Authorized  Standards,  and  in  Forensic  Examinations  for  Poisons, 
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8vo,  cloth $5.00 

Outlines  of  Proximate  Organic  Analysis,  for  the  Iden- 
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and    OTIS    COE    JOHNSON.     Qualitative   Chemical 

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With  an  Appendix  by  H.  H.  Willard,  containing  a  few  improved 
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STANDARD  TEXT  BOOKS.  11 

PROST,  E.     Manual  of  Chemical  Analysis  as  Applied  to 

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Products.  Translated  from  the  original  by  J.  C.  Smith.  Part  I, 
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Part  II,  Metals;  Part  III,  Alloys.  8vo,  cloth net,  $4.50 

RANKINE,  W.  J.  MACQUORN,  C.E.,  LL.D.,  F.R.S.  Ma- 
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Work,  Strength,  Construction,  and  Objects  of  Machines,  etc. 
Illustrated  with  nearly  300  woodcuts.  Seventh  Edition.  Thor- 
oughly revised  by  W.  J.  Millar.  8vo,  cloth $5.00 

The  Steam-Engine  and  Other  Prime  Movers.     With 

diagrams  of  the  Mechanical  Properties  of  Steam.  With  folding 
plates,  numerous  tables  and  illustrations.  Fifteenth  Edition. 
Thoroughly  revised  by  W.  J.  Millar.  8vo,  cloth $5.00 

Useful  Rules  and  Tables  for  Engineers  and  Others. 

With  appendix,  tables,  tests,  and  formulae  for  the  use  of  Electrical 
Engineers.  Comprising  Submarine  Electrical  Engineering,  Electric 
Lighting,  and  Transmission  of  Power.  By  Andrew  Jamieson, 
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Millar.  8vo,  cloth $4.00 

A  Mechanical  Text-Book.     By  Prof.  Macquorn  Ran- 

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Applied    Mechanics.     Comprising    the    Principles    of 

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With  numerous  tables  and  illustrations.  Twenty-first  Edition. 
Thoroughly  revised  by  W.  J.  Millar.  8vo,  cloth $6 . 50 

RATEAU,   A.     Experimental  Researches  on  the   Flow  of 

Steam  Through  Nozzles  and  Orifices,  to  which  is  added  a  note  on 
The  Flow  of  Hot  Water.  Authorized  translation  by  H.  Boyd 
Brydon.  12mo,  cloth,  illustrated ; net,  $1 . 50 

RAUTENSTRAUCH,  W.     Syllabus  of  Lectures  and  Notes 

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REINHARDT,    CHAS.    W.     Lettering    for    Draughtsmen, 

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RIPPER,  WILLIAM.  A  Course  of  Instruction  in  Machine 
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ROBINSON,   S.   W.     Practical  Treatise   on   the   Teeth  of 

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SEIDELL,   A.     Handbook    of    Solubilities.      12mo,    cloth. 

In  Press. 


STANDARD  TEXT  BOOKS.  13 

SEVER,  Prof.  G.  F.     Electrical  Engineering  Experiments 

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SEWALL,   C.  H.     Lessons  in  Telegraphy.     For  use  as  a 

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SHELDON,  Prof.  S.,  Ph.D.,  and  MASON,  HOBART,  B.S. 

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SHIELDS,    J.    E.     Notes    on    Engineering    Construction. 

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SHUNK,    W.    F.     The    Field   Engineer.     A   Handy   Book 

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WRIGHT,  Prof.  T.  W.      Elements  of  Mechanics,  including 

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